Molecular Biology of
THE CELL
Sixth Edition
The Problems Book
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Molecular Biology of
THE CELL
Sixth Edition
The Problems Book
John Wilson and Tim Hunt
Garland Science
Vice President: Denise Schanck
Associate Editor: Allie Bochicchio
Production Editor and Layout: EJ Publishing Services
Senior Production Editor: Georgina Lucas
Master Reviewer: Alastair Ewing
Copyeditor: Jo Clayton
Proofreader: Sally Huish
Illustrator: Nigel Orme
Designer: Matthew McClements, Blink Studio, Ltd.
Back cover photograph and design: Nigel Orme
Indexer: Medical Indexing Ltd.
Permissions Coordinator: Sheri Gilbert
© 2015 by John Wilson and Tim Hunt
John Wilson received his PhD from the California Institute of Technology and
pursued his postdoctoral work at Stanford University. He is currently Distinguished
Service Professor of Biochemistry and Molecular Biology at Baylor College of
Medicine in Houston. His research interests include genome instability and gene
therapy. He has taught medical and graduate students for many years, coauthored
books on immunology, molecular biology, and biochemistry, and received
numerous teaching honors, including the Robertson Presidential Award for
excellence in education.
Tim Hunt received his PhD from the University of Cambridge where he taught
biochemistry and cell biology for more than 20 years. In the late 1970s and early
1980s he spent his summers teaching cell and molecular biology at the Marine
Biological Laboratory, Woods Hole, Massachusetts. He left Cambridge in 1990 and
moved to the Cancer Research UK Clare Hall Laboratories, just outside London,
where he worked on the control of the cell cycle until his retirement in 2010.
He shared the 2001 Nobel Prize in Physiology or Medicine with Lee Hartwell
and Paul Nurse.
This book contains information obtained from authentic and highly regarded
sources. Every effort has been made to trace copyright holders and to obtain their
permission for the use of copyright material. Reprinted material is quoted with
permission, and sources are indicated. A wide variety of references are listed.
Reasonable efforts have been made to publish reliable data and information, but
the author and the publisher cannot assume responsibility for the validity of all
materials or for the consequences of their use.
All rights reserved. No part of this book covered by the copyright herein may be
reproduced or used in any format in any form or by any means—graphic, electronic,
or mechanical, including photocopying, recording, taping, or information storage
and retrieval systems—without permission of the publisher.
ISBN: 978-0-8153-4453-7 (paperback)
Published by Garland Science, Taylor & Francis Group, LLC, an informa business,
711 Third Avenue, New York, NY 10017, US
3 Park Square, Milton Park, Abingdon, OX14 4RN, UK
Printed in the United States of America
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Visit our website at http://www.garlandscience.com
We dedicate this book to the memory of our comrade, Julian Lewis.
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vii
Preface
Welcome to he Problems Book, which aims to help students appreciate the ways
in which an understanding of how cells work, as discussed in Molecular Biology
of the Cell, Sixth Edition, by Alberts et al., can be further explored through experiments and simple calculations. As always, we hope to stimulate our readers to
ask questions as well as to learn and digest the stories that “the big book” tells. In
real life, knowledge and understanding come from research, which entails curiosity, puzzlement, doubt, criticism, and debate as well as performing experiments.
Groping one’s way through the fog of uncertainty during a project is a slow and
often discouraging process; eureka moments (even if one is lucky) are few and far
between. Nevertheless, those moments catch the essence of the drama, and we
have tended to focus on them, where we have been able to cast them in the form
of a problem. In this way, for student and teacher alike, we hope to encourage a
questioning attitude to biology. Without curiosity there would be neither science
nor scientists.
We have been making up problems together since 1985, and the revision leading to this new edition of he Problems Book has taken us more than four years.
here are several new things about this edition. First, the book is now in color. As
well as improving its look as a whole, we think this will improve the clarity and
intelligibility of the igures. Second, we’ve added a new type of question, MCAT
Style, modeled on the kind of problems that are found in most medical school
admissions tests. hese were drafted by Doug Kellogg at the University of California, Santa Cruz, and we think they make a great addition to the book. Indeed,
we were pleasantly surprised to discover that these questions allowed us to frame
problems in new and interesting ways. Elsewhere, we have done a considerable
amount of pruning, partly to make space for these new problems and partly to
eliminate problems that were showing their age or were no longer relevant to the
parent text.
he organization of he Problems Book remains largely the same. here are
Terms to Learn, Deinitions, and True/False sections in every chapter. Next come
hought Problems, of which some are more challenging than others—they may
be playful, or serious, but all are designed to make the reader think. Following
these is a section called Calculations, designed to deal with quantitative aspects
of cell biology. he calculations in this book are mostly very straightforward, usually involving no more than the interconversion of units, yet they provide a solid
framework for thinking about the cell. Are cell-surface receptors sparse in the
plasma membrane, or jam-packed? Do molecules difuse across a cell slowly, or
in the blink of an eye? Does chromatin occupy most of the nuclear volume, or
just a tiny fraction? How fast could a tomato plant grow, theoretically? Numerical analysis of such questions is very important if one is to gain an understanding for the molecular basis of cell biology. he Data Handling section contains
research-based problems. Our original brief was to compose problems based on
experiments so as to allow readers to get a better feel for the way in which biological knowledge is obtained. It is tremendously important to keep asking, “How do
we know that? What’s the evidence?” or to wonder how one might go about inding something out. Often it’s not at all obvious, often the initial breakthrough was
a lucky chance observation, made while investigating some completely diferent
viii
PREFACE
business. In fact, it takes most of us years of research experience to grasp the idea
of how one simple fact “can illuminate a distant area, hitherto dark” (Boveri,
1902). Seeing how these tiny shards of evidence give rise to the big picture often
involves considerable imagination as well as a certain discipline, to know how
much weight the evidence will bear. We hope we have sometimes, at least, been
able to capture the essence of how experiments lead to understanding. To do
justice to the authors of the experiments we use in these problems, however, we
strongly recommend recourse to the original papers, whose references we always
provide.
A newly compiled section, Medical Links, contains problems of particular
interest to health science students.
We hope that the organization and classiication of problems will help both
student and teacher to ind what they are looking for. How should this book be
used? We composed it by a process of constant dialogue and discussion, and we
suspect that the most fruitful use of the problems will be to stimulate discussions
in class, or between students. Tackling selected problems as homework will also
surely help. Teachers have told us that they ind ideas for exam questions here,
and all the answers to our questions are now provided in he Problems Book, for
many of these problems are diicult to answer and are not intended to be set as
tests. Rather, we hope that readers will be intrigued (as we were) by the questions
we ask, and after thinking a bit will want to see what the answer is, what form the
discussion takes, and how to get at thinking about this particular kind of a problem.
he answers to the end-of-chapter problems in Molecular Biology of the Cell,
Sixth Edition can also be found in the back of this book, including the answers to
newly written problems for Chapters 21 through 24.
As always, we want to hear from our readers, for despite our best eforts, we do
not always get things right. Please email John Wilson at jwilson@bcm.edu or Tim
Hunt at tim.hunt@cancer.org.uk with your comments or queries, and we’ll do our
best to answer them.
ix
Acknowledgments
It remains true that our rate of production, over the years, averages at about one
chapter per year, but even this glacial progress would not have been possible
without a tremendous amount of help from friends and colleagues whose names
are recorded in previous editions of he Problems Book, which appeared in 1989,
1994, 2002, and 2008. As ever, we are indebted to Alastair Ewing who worked
through all the new problems, discovering embarrassing mistakes and inding
better, clearer, and more graceful ways of putting things. Denise Schanck has been
a tower of strength, as always, and Emma Jefcock a brilliant and startlingly eicient designer and friend. Allie Bochicchio coordinated our activities and kept us
in good order. Mike Morales helped to set up an instant home-away-from-home
during meetings in California, and his cheerful humor helped tremendously.
Adam Sendrof, who took care of marketing and gave us audience feedback, was
unfailingly supportive. We are especially grateful to all the authors of Molecular
Biology of the Cell, who have been extremely helpful in the selection and reinement of problems that appear in the main text. We thank them most warmly for
their suggestions. Once again, Nigel Orme has been a huge help with the illustrations, particularly in the business of adding color. he person who made the most
important contribution to this edition, however, is Doug Kellogg of the University of California, Santa Cruz. Doug is an old friend who allowed himself to be
persuaded to take on the task of composing MCAT-style questions, despite his
heavy responsibilities as a teacher, researcher, and new father. We are very lucky,
however, to have the loving support of our spouses and families, Lynda, Mary,
Celia, and Aggie. hey’ve put up heroically with our regular absences and preoccupations.
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xi
A Couple of Useful Things
to Know
Avogadro’s Number (6.02 × 1023 molecules/mole)
Avogadro’s number (N) is perhaps the most important constant in molecular sciences,
and it appears again and again in this book. Do you know how it was determined? We
didn’t, or had forgotten if we ever knew. How can one measure the number of molecules in a mole? And who did it irst? You will not ind this information in modern
biology books, partly because it is ancient history, and partly because it was the business of physicists; some pretty good physicists too, as we shall see.
Amadeo Avogadro had no idea how many molecules there were in 22.4 L of a gas.
His hypothesis, presented in 1811, was simply that equal volumes of all gases contained the same number of molecules, irrespective of their size or density. Not until
much later, when the reality of molecules was more widely accepted and the microscopic basis for the properties of gases was being worked out, were the irst estimates
attempted. An Austrian high school teacher called Josef Loschmidt used James Clerk
Maxwell’s recently developed kinetic theory of gases to estimate how many molecules
there were in a cubic centimeter of air. Maxwell had derived an expression for the viscosity of a gas, which is proportional to the density of the gas, to the mean velocity of
the molecules, and to their mean free path. he latter could be estimated if one knew
the size and number of the molecules. Loschmidt simply made the assumption that
when a gas was condensed into a liquid, its molecules were packed as closely as they
could be, like oranges in a display on a fruit stand, and from this he was able to get a
pretty accurate value for Avogadro’s number. Not surprisingly, in Austria they often
refer to N as “Loschmidt’s number.” In fact, it wasn’t until 1909 that the term “Avogadro’s number” was suggested by Jean Perrin, who won the 1926 Nobel Prize in Physics
(his lecture is available on the Nobel web site, and his book entitled, simply, Atoms
[Les Atomes, 1913, translated from the original French by D.L. Hammick, reprinted in
1990 by Ox Bow Press] is highly recommended—and accessible—reading. It has been
called the inest book on physics of the twentieth century).
You may be surprised to discover, as we were, that estimating Avogadro’s number
was an important component of Albert Einstein’s Ph.D. thesis. Abraham Pais’s wonderful biography of Einstein, Subtle is the Lord (subtitled he Science and the Life of
Albert Einstein, 1982, Oxford University Press), devotes Chapter 5, “he Reality of Molecules,” to this period of the great physicist’s life and work. Einstein found three independent ways to estimate N: from the viscosity of dilute sucrose solutions, from his
analysis of Brownian motion, and from light-scattering by gases near the critical point,
including the blueness of the sky. Because the sky is ive million times less bright than
direct sunlight, Avogadro’s number is 6 × 1023. Isn’t that romantic?
But Einstein’s was not the last word on the subject. Indeed, according to Pais, he
made an “elementary but nontrivial mistake” in his thesis that was later corrected,
and it was really Perrin who brought the whole ield together with his experiments on
Brownian motion. he Nobel presentation speech contains this line:
“His [Perrin’s] measurements on the Brownian movement showed that Einstein’s
theory was in perfect agreement with reality. hrough these measurements a new
determination of Avogadro’s number was obtained.”
For most methods of counting molecules, neither the physics nor the math is easy
to follow, but two are simple to understand. he irst comes from radioactive decay,
and another Nobel prize-winning physicist, Ernest Rutherford. When radium decays,
it emits alpha particles, which are helium nuclei. If you can count the radioactive
decay events with a Geiger counter and measure the volume of helium emitted, you
can estimate Avogadro’s number. he second way is much more modern. You can see
large proteins and nucleic acids with the aid of an electron microscope, and count
them directly!
xii
A Couple of Things to Know
Calculations and Unit Analysis
Many of the problems in this book involve calculations. Where the calculations are
based on an equation (for example, the Nernst equation or the equation for the volume of a sphere), we provide the equation along with a brief explanation of symbols,
and often their values. Many calculations, however, involve the conversion of information from one form into another, equivalent form. For example, if the concentration of a protein is 10–9 M, how many molecules of it would be present in a mammalian nucleus with a volume of 500 μm3? Here, a concentration is given as M (moles/L),
whereas the desired answer is molecules/nucleus; both values are expressed as
“number/volume” and the problem is to convert one into the other.
Both kinds of calculation use constants and conversion factors that may or may
not be included in the problem. he Nernst equation, for example, uses the gas constant R (8.3 × 10–3 kJ/K mole) and the Faraday constant F (96 kJ/V mole). And conversion of moles/L to molecules/nucleus requires Avogadro’s number N (6.0 × 1023 molecules/mole). All of the constants, symbols, and conversion factors that are used in
this book are listed in Tables 1–8 at the end of the book, on pages 963–966 (including
the standard genetic code, the one-letter amino acid code, useful geometric formulas,
and data on common radioisotopes used in biology).
For each type of calculation, we strongly recommend the powerful general strategy
known as unit analysis (or dimensional analysis). If units (for example, moles/L) are
included along with the numbers in the calculations, they provide an internal check
on whether the numbers have been combined correctly. If you’ve made a mistake in
your math, the units will not help, but if you’ve divided where you should have multiplied, for example, the units of the answer will be nonsensical: they will shout “error.”
Consider the conversion of 10–9 M (moles/L) to molecules/nucleus. In the conversion
of moles to molecules, do you multiply 10–9 by 6 × 1023 (Avogadro’s number) or do you
divide by it? If units are included, the answer is clear.
10–9 moles
L
10–9 moles
L
×
×
6 × 1023 molecules
6 × 1014 molecules
=
mole
L
mole
6 × 1023 molecules
=
1.7 × 10–33 mole2
YES
NO
molecules L
Similarly, in the conversion of liters to nuclei, the goal is to organize the conversion factors to transform the units to the desired form.
300 molecules
1L
mL
6 × 1014 molecules
cm3
500 µm3
×
×
×
=
×
4
3
3
nucleus
nucleus
1000 mL
L
(10 µm)
cm
If you do this calculation with pure numbers, you must worry at each step whether
to divide or multiply. If you attach the units, however, the decision is obvious. It is
important to realize that any set of (correct) conversion factors will give the same
answer. If you are more comfortable converting liters to ounces, that’s ine, so long as
you know a string of conversion factors that will ultimately transform ounces to μm3.
here are a few simple rules for handling units in calculations.
1. Quantities with diferent units cannot be added or subtracted. (You cannot
subtract 3 meters from 10 kJ.)
2. Quantities with diferent units can be multiplied or divided; just multiply or
divide the units along with the numbers. (You can multiply 3 meters times 10
kJ; the answer is 30 kJ meters.)
3. All exponents are unitless. (You can’t use 106 mL.)
4. You cannot take the logarithm of a quantity with units.
hroughout this book, we have included the units for each element in every calculation. If the units are arranged so that they cancel to give the correct units for the
answer, the numbers will take care of themselves.
xiii
Contents
Problems
Chapter 1
Cells and Genomes
1
Chapter 2
Cell Chemistry and Bioenergetics
11
Chapter 3
Proteins
31
Chapter 4
DNA, Chromosomes, and Genomes
53
Chapter 5
DNA Replication, Repair, and Recombination
77
Chapter 6
How Cells Read the Genome: From DNA
to Protein
105
Chapter 7
Control of Gene Expression
135
Chapter 8
Analyzing Cells, Molecules, and Systems
167
Chapter 9
Visualizing Cells
197
Chapter 10
Membrane Structure
209
Chapter 11
Membrane Transport of Small Molecules and the
Electrical Properties of Membranes
219
Chapter 12
Intracellular Compartments and Protein Sorting
237
Chapter 13
Intracellular Membrane Trafic
259
Chapter 14
Energy Conversion: Mitochondria and
Chloroplasts
285
Chapter 15
Cell Signaling
307
Chapter 16
The Cytoskeleton
333
Chapter 17
The Cell Cycle
361
Chapter 18
Cell Death
387
Chapter 19
Cell Junctions and the Extracellular Matrix
393
Chapter 20
Cancer
415
Answers to Problems
435
Answers to Problems in Molecular Biology of the
Cell, Sixth Edition
861
Credits
931
Index
935
Tables
963
xiv
Detailed Contents
Problems
Chapter 1
Cells and Genomes
1
HoMoloGoUS RECoMBINATIoN
95
TRANSPoSITIoN AND CoNSERVATIVE
SITE-SPECIFIC RECoMBINATIoN
99
4
Chapter 6
How Cells Read the Genome:
From DNA to Protein
105
6
FRoM DNA To RNA
105
FRoM RNA To PRoTEIN
118
THE UNIVERSAl FEATURES oF CEllS oN EARTH
1
THE DIVERSITy oF GENoMES AND THE TREE oF lIFE
GENETIC INFoRMATIoN IN EUkARyoTES
Chapter 2
Cell Chemistry and
Bioenergetics
11
THE RNA WoRlD AND THE oRIGINS oF lIFE
130
THE CHEMICAl CoMPoNENTS oF A CEll
11
Chapter 7
135
CATAlySIS AND THE USE oF ENERGy By CEllS
19
HoW CEllS oBTAIN ENERGy FRoM FooD
Control of Gene Expression
AN oVERVIEW oF GENE CoNTRol
135
24
CoNTRol oF TRANSCRIPTIoN By
SEQUENCE-SPECIFIC DNA-BINDING PRoTEINS
137
31
TRANSCRIPTIoN REGUlAToRS SWITCH GENES
oN AND oFF
144
THE SHAPE AND STRUCTURE oF PRoTEINS
31
MolECUlAR GENETIC MECHANISMS THAT CREATE
AND MAINTAIN SPECIAlIZED CEll TyPES
149
PRoTEIN FUNCTIoN
37
MECHANISMS THAT REINFoRCE CEll MEMoRy
IN PlANTS AND ANIMAlS
154
PoST-TRANSCRIPTIoNAl CoNTRolS
157
REGUlATIoN oF GENE EXPRESSIoN By
NoNCoDING RNAS
161
Chapter 8
Analyzing Cells, Molecules,
and Systems
167
167
168
Chapter 3
Chapter 4
Genomes
Proteins
DNA, Chromosomes, and
53
THE STRUCTURE AND FUNCTIoN oF DNA
53
CHRoMoSoMAl DNA AND ITS PACkAGING
IN THE CHRoMATIN FIBER
55
CHRoMATIN STRUCTURE AND FUNCTIoN
59
THE GloBAl STRUCTURE oF CHRoMoSoMES
64
ISolATING CEllS AND GRoWING THEM IN
CUlTURE
67
PURIFyING PRoTEINS
HoW GENoMES EVolVE
Chapter 5
DNA Replication, Repair,
and Recombination
77
THE MAINTENANCE oF DNA SEQUENCES
77
DNA REPlICATIoN MECHANISMS
78
THE INITIATIoN AND CoMPlETIoN oF DNA
REPlICATIoN IN CHRoMoSoMES
84
DNA REPAIR
90
ANAlyZING PRoTEINS
172
ANAlyZING AND MANIPUlATING DNA
177
STUDyING GENE EXPRESSIoN AND FUNCTIoN
184
MATHEMATICAl ANAlySIS oF CEll FUNCTIoNS
190
Chapter 9
Visualizing Cells
197
lookING AT CEllS IN THE lIGHT MICRoSCoPE
197
lookING AT CEllS AND MolECUlES IN THE
ElECTRoN MICRoSCoPE
204
Detailed Contents
Chapter 10
Membrane Structure
xv
209
THE lIPID BIlAyER
209
MEMBRANE PRoTEINS
214
THE GENETIC SySTEMS oF MIToCHoNDRIA
AND CHloRoPlASTS
301
Chapter 15
307
Cell Signaling
PRINCIPlES oF CEll SIGNAlING
307
SIGNAlING THRoUGH G-PRoTEIN-CoUPlED
RECEPToRS
312
219
SIGNAlING THRoUGH ENZyME-CoUPlED
RECEPToRS
320
TRANSPoRTERS AND ACTIVE MEMBRANE
TRANSPoRT
222
AlTERNATIVE SIGNAlING RoUTES IN GENE
REGUlATIoN
325
CHANNElS AND THE ElECTRICAl PRoPERTIES
oF MEMBRANES
227
SIGNAlING IN PlANTS
329
Chapter 16
The Cytoskeleton
333
FUNCTIoN AND oRIGIN oF THE CyToSkElEToN
333
ACTIN AND ACTIN-BINDING PRoTEINS
335
MyoSIN AND ACTIN
340
MICRoTUBUlES
344
Chapter 11 Membrane Transport of
Small Molecules and the Electrical
Properties of Membranes
219
PRINCIPlES oF MEMBRANE TRANSPoRT
Chapter 12 Intracellular Compartments
and Protein Sorting
237
THE CoMPARTMENTAlIZATIoN oF CEllS
237
THE TRANSPoRT oF MolECUlES BETWEEN THE
NUClEUS AND THE CyToSol
240
THE TRANSPoRT oF PRoTEINS INTo
MIToCHoNDRIA AND CHloRoPlASTS
246
INTERMEDIATE FIlAMENTS AND SEPTINS
353
PERoXISoMES
249
CEll PolARIZATIoN AND MIGRATIoN
356
THE ENDoPlASMIC RETICUlUM
252
Chapter 17
The Cell Cycle
361
oVERVIEW oF THE CEll CyClE
361
THE CEll-CyClE CoNTRol SySTEM
364
S PHASE
366
MIToSIS
368
CyTokINESIS
376
MEIoSIS
379
CoNTRol oF CEll DIVISIoN AND CEll GRoWTH
380
Chapter 18
Cell Death
387
Chapter 13
Trafic
Intracellular Membrane
THE MolECUlAR MECHANISMS oF MEMBRANE
TRANSPoRT AND THE MAINTENANCE oF
CoMPARTMENTAl DIVERSITy
259
259
TRANSPoRT FRoM THE ER THRoUGH THE
GolGI APPARATUS
264
TRANSPoRT FRoM THE TRANS GolGI NETWoRk
To lySoSoMES
269
TRANSPoRT INTo THE CEll FRoM THE PlASMA
MEMBRANE: ENDoCyToSIS
272
Chapter 19 Cell Junctions and the
Extracellular Matrix
393
TRANSPoRT FRoM THE TRANS GolGI NETWoRk
To THE CEll EXTERIoR: EXoCyToSIS
277
CEll–CEll JUNCTIoNS
393
THE EXTRACEllUlAR MATRIX oF ANIMAlS
403
CEll–MATRIX JUNCTIoNS
408
THE PlANT CEll WAll
410
Chapter 20
415
Chapter 14 Energy Conversion:
Mitochondria and Chloroplasts
285
THE MIToCHoNDRIoN
285
THE PRoToN PUMPS oF THE ElECTRoNTRANSPoRT CHAIN
287
ATP PRoDUCTIoN IN MIToCHoNDRIA
292
CHloRoPlASTS AND PHoToSyNTHESIS
296
Cancer
CANCER AS A MICRoEVolUTIoNARy PRoCESS
415
CANCER-CRITICAl GENES: HoW THEy ARE
FoUND AND WHAT THEy Do
418
xvi
Detailed Contents
CANCER PREVENTIoN AND TREATMENT:
PRESENT AND FUTURE
426
Answers to Problems
435
Answers to Problems in Molecular Biology
of the Cell, Sixth Edition
861
Credits
931
Index
935
Table 1 Constants
Table 2 Variables
Table 3 Units
Table 4 Preixes
Table 5 Geometric Formulas
Table 6 Radioactive Isotopes
Table 7 The Genetic Code
Table 8 Amino Acids and Codons
963
963
964
965
965
965
966
966
Problems
A Sea Creature Sighted Between
Antibes and Nice in 1562.
We shall never know exactly what the
person who drew this picture actually
saw, but it is doubtful if he (or she)
made it up completely. Anyone who
has taken a histology course will know
how very dificult it is to learn to see and
abstract salient, accurate details from
a completely unfamiliar scene. This is
important for cell biologists; it is natural
to interpret the unfamiliar in terms of
what you already know, blinding you to
the more truthful new. This theme runs
through the opening pictures in this book,
but if you are curious as to what the “sea
creature” really looked like, turn to the
Answers section.
Chapter 1
1
CHAPTER
1
Cells and Genomes
THE UNIVERSAL FEATURES OF CELLS ON EARTH
TERMS TO LEARN
amino acid
DNA replication
enzyme
gene
genome
messenger RNA (mRNA)
nucleotide
plasma membrane
polypeptide
protein
ribonucleic acid (RNA)
transcription
translation
DEFINITIONS
Match each deinition below with its term from the list above.
1–1
he selective barrier surrounding a living cell that enables the cell to concentrate nutrients, retain products, and excrete waste.
1–2
A protein that catalyzes a speciic chemical reaction.
1–3
he copying of one strand of DNA into a complementary RNA sequence.
1–4
Process by which the sequence of nucleotides in an mRNA molecule
directs the incorporation of amino acids into protein.
1–5
Region of DNA that controls a discrete hereditary characteristic of an
organism, usually corresponding to a single protein (or set of alternative
protein variants) or to a structural, catalytic, or regulatory RNA.
1–6
RNA molecule that speciies the amino acid sequence of a protein.
1–7
he building blocks of proteins.
1–8
he total genetic information of a cell or organism as embodied in its
complete DNA sequence.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
1–9
Genes and their encoded proteins are co-linear; that is, the order of
amino acids in proteins is the same as the order of the codons in the RNA
and DNA.
1–10
DNA and RNA use the same four-letter alphabet.
THOUGHT PROBLEMS
1–11
“Life” is easy to recognize but diicult to deine. he dictionary deines
life as “he state or quality that distinguishes living beings or organisms
from dead ones and from inorganic matter, characterized chiely by
metabolism, growth, and the ability to reproduce and respond to stimuli.”
IN THIS CHAPTER
THE UNIVERSAL FEATURES OF
CELLS ON EARTH
THE DIVERSITY OF GENOMES
AND THE TREE OF LIFE
GENETIC INFORMATION IN
EUKARYOTES
2
Chapter 1: Cells and Genomes
Biology textbooks usually elaborate slightly; for example, according to a
popular text, living things
1. Are highly organized compared with natural inanimate objects.
2. Display homeostasis, maintaining a relatively constant internal
environment.
3. Reproduce themselves.
4. Grow and develop from simple beginnings.
5. Take energy and matter from the environment and transform it.
6. Respond to stimuli.
7. Show adaptation to their environment.
Score a car, a cactus, and yourself with respect to these seven characteristics.
1–12
NASA has asked you to design a module that will identify signs of life on
Mars. What will your module look for?
1–13
You have embarked on an ambitious research project: to create life in a
test tube. You boil up a rich mixture of yeast extract and amino acids in a
lask along with a sprinkling of the inorganic salts known to be essential
for life. You seal the lask and allow it to cool. After several months, the
liquid is as clear as ever, and there are no signs of life. A friend suggests
that excluding air was a mistake, since most life as we know it requires
oxygen. You repeat the experiment, but this time you leave the lask open
to the atmosphere. To your great delight, the liquid becomes cloudy after
a few days and under the microscope you see beautiful small cells that
are clearly growing and dividing. Does this experiment prove that you
managed to generate a novel life-form? How might you redesign your
experiment to allow air into the lask, yet eliminate the possibility that
contamination is the explanation for the results?
1–14
he genetic code (see Tables 7 and 8, page 966) speciies the entire set of
codons that relate the nucleotide sequence of mRNA to the amino acid
sequence of encoded proteins. Since the code was deciphered nearly
four decades ago, some have claimed that it must be a frozen accident,
while others have argued that it was shaped by natural selection.
A striking feature of the genetic code is its inherent resistance to the
efects of mutation. For example, a change in the third position of a codon
often speciies the same amino acid or one with similar chemical properties. But is the natural code more resistant to mutation (less susceptible
to error) than other possible versions? he answer is an emphatic “Yes,”
as illustrated in Figure 1–1. Only one in a million computer-generated
“random” codes is more error-resistant than the natural genetic code.
Does the extraordinary mutation resistance of the genetic code argue
in favor of its origin as a frozen accident or as a result of natural selection?
Explain your reasoning.
number of codes (thousands)
25
20
15
10
natural
code
5
0
0
5
10
15
susceptibility to mutation
20
Figure 1–1 Susceptibility to mutation of
the natural code shown relative to that
of millions of other computer-generated
alternative genetic codes (Problem 1–14).
Susceptibility measures the average
change in amino acid properties caused
by random mutations in a genetic code.
A small value indicates that mutations
tend to cause only minor changes.
THE UNIVERSAL FEATURES OF CELLS ON EARTH
1–15
You have begun to characterize a sample obtained from the depths of
the oceans on Europa, one of Jupiter’s moons. Much to your surprise,
the sample contains a life-form that grows well in a rich broth. Your preliminary analysis shows that it is cellular and contains DNA, RNA, and
protein. When you show your results to a colleague, she suggests that
your sample was contaminated with an organism from Earth. What
approaches might you try to distinguish between contamination and a
novel cellular life-form based on DNA, RNA, and protein?
1–16
In the 1940s, Erwin Chargaf made the remarkable observation that in
samples of DNA from a wide range of organisms the mole percent of G
[G/(A+T+C+G)] was equal to the mole percent of C, and the mole percents of A and T were equal. his was an essential clue to the structure of
DNA. Nevertheless, Chargaf’s “rules” were not universal. For example,
in DNA from the virus ΦX174, which has a single-stranded genome, the
mole percents are A = 24, C = 22, G = 23, and T = 31. What is the structural
basis for Chargaf’s rules, and how is it that DNA from ΦX174 doesn’t
obey the rules?
1–17
In 1944, at the beginning of his book What is Life, the great physicist Erwin
SchrÖdinger (of cat fame) asked the following question: “How can the
events in time and space which take place within the spatial boundary
of a living organism be accounted for by physics and chemistry?” What
would be your answer today? Do you think there are peculiar properties
of living systems that disobey the laws of physics and chemistry?
1–18
Which of the following correctly describe the coding relationships (template → product) for replication, transcription, and translation?
A. DNA → DNA
B. DNA → RNA
C. DNA → protein
D. RNA → DNA
E. RNA → RNA
F. RNA → protein
G. Protein → DNA
H. Protein → RNA
I. Protein → protein
CALCULATIONS
1–19
An adult human is composed of about 1013 cells, all of which are derived
by cell division from a single fertilized egg.
A. Assuming that all cells continue to divide (like bacteria in rich media),
how many generations of cell divisions would be required to produce
1013 cells?
B. Human cells in culture divide about once per day. Assuming that all cells
continue to divide at this rate during development, how long would it
take to generate an adult organism?
C. Why is it, do you think, that adult humans take longer to develop than
these calculations might suggest?
1–20
here are 21,000 protein-coding genes in the human genome. If you
wanted to use a stretch of the DNA of each gene as a unique identiication
tag, roughly what minimum length of DNA sequence would you need? To
be unique, the length of DNA in nucleotides would have to have a diversity (the number of diferent possible sequences) equivalent to at least
21,000 and would have to be present once in the haploid human genome
(3.2 × 109 nucleotides). (Assume that A, T, C, and G are present in equal
amounts in the human genome.)
1–21
Cell growth depends on nutrient uptake and waste disposal. You might
imagine, therefore, that the rate of movement of nutrients and waste
3
4
Chapter 1: Cells and Genomes
products across the cell membrane would be an important determinant
of the rate of cell growth. Is there a correlation between a cell’s growth
rate and its surface-to-volume ratio? Assuming that the cells are spheres,
compare a bacterium (radius 1 μm), which divides every 20 minutes,
with a human cell (radius 10 μm), which divides every 24 hours. Is there
a match between the surface-to-volume ratios and the doubling times for
these cells? [he surface area of a sphere = 4πr2; the volume = (4/3)πr3.]
THE DIVERSITY OF GENOMES AND THE TREE OF LIFE
TERMS TO LEARN
archaea
bacteria
eukaryote
gene family
homolog
model organism
mutation
ortholog
paralog
prokaryote
virus
DEFINITIONS
Match each deinition below with its term from the list above.
1–22
A small packet of genetic material that has evolved as a parasite on the
reproductive and biosynthetic machinery of host cells.
1–23
Organism selected for intensive study as a representative of a large group
of species.
1–24
One of the two divisions of prokaryotes, typically found in hostile environments such as hot springs or concentrated brine.
1–25
he general term for genes that are related by descent.
1–26
Living organism composed of one or more cells with a distinct nucleus
and cytoplasm.
1–27
Major category of living cells distinguished by the absence of a nucleus.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
1–28
he vast majority of CO2 ixation into the organic compounds needed for
further biosynthesis is carried out by phototrophs.
1–29
Each member of the human hemoglobin gene family, which consists of
seven genes arranged in two clusters on diferent chromosomes, is an
ortholog to all of the other members.
THOUGHT PROBLEMS
1–30
It is not so diicult to imagine what it means to feed on the organic molecules that living things produce. hat is, after all, what we do. But what
does it mean to “feed” on sunlight, as phototrophs do? Or, even stranger,
to “feed” on rocks, as lithotrophs do? Where is the “food,” for example, in
the mixture of chemicals (H2S, H2, CO, Mn+, Fe2+, Ni2+, CH4, and NH4+)
spewed forth from a hydrothermal vent?
1–31
At the bottom of the seas where hydrothermal vents pour their chemicals into the ocean, there is no light and little oxygen, yet giant (2-meter
long) tube worms live there happily. hese remarkable creatures have no
mouth and no anus, living instead of the excretory products and dead
cells of their symbiotic lithotrophic bacteria. hese tube worms are bright
red because they contain large amounts of hemoglobin, which is critical
to the survival of their symbiotic bacteria, and, hence, the worms. his
THE DIVERSITY OF GENOMES AND THE TREE OF LIFE
5
specialized hemoglobin carries O2 and H2S. In addition to providing O2
for the tube worm’s oxidative metabolism, what role might this specialized hemoglobin play in the symbiotic relationship that is crucial for life
in this hostile environment?
1–32
he overall reaction for the production of glucose (C6H12O6) by oxygenic
(oxygen-generating) photosynthesis,
6 CO2 + 6 H2O + light → C6H12O6 + 6 O2
(Equation 1)
was widely interpreted as meaning that light split CO2 to generate O2,
and that the carbon was joined with water to generate glucose. In the
1930s, a graduate student at Stanford University, C.B. van Neil, showed
that the stoichiometry for photosynthesis by purple sulfur bacteria was
6 CO2 + 12 H2S + light → C6H12O6 + 6 H2O + 12 S
(Equation 2)
On the basis of this stoichiometry, he suggested that the oxygen generated during oxygenic photosynthesis derived from water, not CO2. His
hypothesis was conirmed two decades later using isotopically labeled
water. Yet how is it that the 6 H2O in Equation 1 can give rise to 6 O2? Can
you suggest how Equation 1 might be modiied to clarify exactly how the
products are derived from the reactants?
1–33
How many possible diferent trees (branching patterns) can in theory
be drawn to display the evolution of bacteria, archaea, and eukaryotes,
assuming that they all arose from a common ancestor?
1–34
he genes for ribosomal RNA are highly conserved (relatively few
sequence changes) in all organisms on Earth; thus, they have evolved
very slowly over time. Were such genes “born” perfect?
1–35
Several prokaryotic genomes have been completely sequenced and their
protein-coding genes have been counted. But how do you suppose one
recognizes a gene in a string of Ts, As, Cs, and Gs?
1–36
Which one of the processes listed below is NOT thought to contribute
signiicantly to the evolution of new genes? Why not?
A. Duplication of genes to create extra copies that can acquire new functions
B. Formation of new genes de novo from noncoding DNA in the genome
C. Horizontal transfer of DNA between cells of diferent species
D. Mutation of existing genes to create new functions
E. Shuling of domains of genes by gene rearrangement
1–37
Genes participating in informational processes such as replication, transcription, and translation are transferred between species much less
often than are genes involved in metabolism. he basis for this inequality
is unclear at present, but one suggestion is that it relates to the underlying
complexity of the two types of processes. Informational processes tend to
involve large aggregates of diferent gene products, whereas metabolic
reactions are usually catalyzed by enzymes composed of a single protein.
A. Archaea are more closely related to bacteria in their metabolic genes, but
are more similar to eukaryotes in the genes involved in informational
processes. In terms of evolutionary descent, do you think archaea separated more recently from bacteria or eukaryotes?
B. Why would the complexity of the underlying process—informational or
metabolic—have any efect on the rate of horizontal gene transfer?
1–38
Why do you suppose that horizontal gene transfer is more prevalent in
single-celled organisms than in multicellular organisms?
1–39
You are interested in inding out the function of a particular gene in the
mouse genome. You have sequenced the gene, deined the portion that
6
Chapter 1: Cells and Genomes
codes for its protein product, and searched the appropriate databases;
however, neither the gene nor the encoded protein resembles anything
seen before. What types of information about the gene or the encoded
protein would you like to know in order to narrow down the possible
functions, and why? Focus on the information you want, rather than on
the techniques you might use to get that information.
CALCULATIONS
1–40
Natural selection is such a powerful force in evolution because cells with
even a small growth advantage quickly outgrow their competitors. To
illustrate this process, consider a cell culture that initially contains 106
bacterial cells, which divide every 20 minutes. A single cell in this culture acquires a mutation that allows it to divide with a generation time
of only 15 minutes. Assuming that there is an unlimited food supply and
no cell death, how long would it take before the progeny of the mutated
cell became predominant in the culture? he number of cells N in the
culture at time t is described by the equation N = N0 × 2t/G, where N0 is the
number of cells at zero time and G is the generation time. (Before you go
through the calculation, make a guess: do you think it would take about a
day, a week, a month, or a year?)
GENETIC INFORMATION IN EUKARYOTES
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
1–41
Eukaryotic cells contain either mitochondria or chloroplasts, but not
both.
1–42
Most of the DNA sequences in a bacterial genome code for proteins,
whereas most of the DNA sequences in the human genome do not.
1–43
he only horizontal gene transfer that has occurred in animals is from the
mitochondrial genome to the nuclear genome.
THOUGHT PROBLEMS
1–44
Animal cells have neither cell walls nor chloroplasts, whereas plant cells
have both. Fungal cells are somewhere in between; they have cell walls
but lack chloroplasts. Are fungal cells more likely to be animal cells that
gained the ability to make cell walls, or plant cells that lost their chloroplasts? his question represented a diicult issue for early investigators who sought to assign evolutionary relationships based solely on cell
characteristics and morphology. How do you suppose that this question
was eventually decided?
1–45
Giardiasis is an acute form of gastroenteritis caused by the protozoan
parasite Giardia lamblia. Giardia is a fascinating eukaryote; it contains
a nucleus but no mitochondria and no recognizable endoplasmic reticulum or Golgi apparatus—one of the very rare examples of such a cellular organization among eukaryotes. his organization might have arisen
because Giardia is an ancient lineage that separated from the rest of
eukaryotes before mitochondria were acquired and internal membranes
were developed. Or it might be a stripped-down version of a more standard eukaryote that has lost these structures because they are not necessary in the parasitic lifestyle it has adopted. How might you use nucleotide sequence comparisons to distinguish between these alternatives?
1–46
Rates of evolution appear to vary in diferent lineages. For example,
the rate of evolution in the rat lineage is signiicantly higher than in the
GENETIC INFORMATION IN EUKARYOTES
7
human lineage. hese rate diferences are apparent whether one looks at
changes in nucleotide sequences that encode proteins and are subject
to selective pressure or at changes in noncoding nucleotide sequences,
which are not under obvious selection pressure. Can you ofer one or
more possible explanations for the slower rate of evolutionary change in
the human lineage versus the rat lineage?
DATA HANDLING
1–47
It is diicult to obtain information about the process of gene transfer
from the mitochondrial to the nuclear genome in animals because there
are few diferences among their mitochondrial genomes. he same set of
13 (or occasionally 12) protein genes is encoded in all the numerous animal mitochondrial genomes that have been sequenced. In plants, however, the situation is diferent, with quite a bit more variability in the sets
of proteins encoded in mitochondrial genomes. Analysis of plants can
thus provide valuable information on the process of gene transfer.
he respiratory gene Cox2, which encodes subunit 2 of cytochrome
oxidase, was functionally transferred to the nucleus during lowering
plant evolution. Extensive analyses of plant genera have pinpointed
the time of appearance of the nuclear form of the gene and identiied
several likely intermediates in the ultimate loss from the mitochondrial
genome. A summary of Cox2 gene distributions between mitochondria
and nuclei, along with data on their transcription, is shown in a phylogenetic context in Figure 1–2.
A. Assuming that transfer of the mitochondrial gene to the nucleus occurred
only once (an assumption supported by the structures of the nuclear
genes), indicate the point in the phylogenetic tree where the transfer
occurred.
B. Are there any examples of genera in which the transferred gene and the
mitochondrial gene both appear functional? Indicate them.
C. What is the minimum number of times that the mitochondrial gene has
been inactivated or lost? Indicate those events on the phylogenetic tree.
GENE
RNA
mt nuc mt nuc
Pisum
+
–
+
–
Clitoria
+
–
+
–
Tephrosia
Galactia
Canavalia
+
+
+
–
–
–
+
+
+
–
–
–
Lespedeza
+
+
+
+
Eriosema
Atylosia
Erythrina
–
–
–
+
+
+
–
–
–
+
+
+
Ramirezella
Vigna
Phaseolus
–
–
–
+
+
+
–
–
–
+
+
+
Dumasia
+
+
+
+
Calopogonium +
Pachyrhizus
+
+
+
+
+
–
–
Cologania
Pueraria
Pseudeminia
Pseudovigna
+
+
+
+
–
–
+
+
+
+
+
+
–
–
+
–
Ortholobium
Psoralea
Cullen
–
–
–
+
+
+
–
–
–
+
+
+
Glycine
+
+
–
+
Neonotonia
Teramnus
Amphicarpa
+
+
+
+
–
+
+
+
+
+
–
+
Figure 1–2 Summary of Cox2 gene
distribution and transcript data in a
phylogenetic context (Problem 1–47).
The presence of the intact gene or the
functional transcript (RNA) is indicated by
(+); the absence of the intact gene or the
functional transcript is indicated by (–).
mt, mitochondria; nuc, nuclei.
8
Chapter 1: Cells and Genomes
D. What is the minimum number of times that the nuclear gene has been
inactivated or lost? Indicate those events on the phylogenetic tree.
E. Based on this information, propose a general scheme for transfer of
mitochondrial genes to the nuclear genome.
1–48
Although stages in the process of mitochondrial gene transfer can be
deduced from studies such as the one in the previous question, there is
much less information on the mechanism by which the gene is transferred from mitochondria to the nucleus. Does a fragment of DNA escape
the mitochondria and enter the nucleus? Or does the transfer somehow
involve an RNA transcript of the gene as the intermediary? he Cox2 gene
provides a unique window on this question. In some species, it is found
in the mitochondrial genome; in others, in the nuclear genome. It so happens that the initial transcript of the mitochondrial Cox2 gene is modiied by RNA editing, a process that changes several speciic cytosines
to uracils. How might this observation allow you to decide whether the
informational intermediary in transfer was DNA or RNA? What do you
think the answer is?
1–49
Some genes evolve rapidly, whereas others are highly conserved. But
how can we tell whether a gene has evolved rapidly or has simply had a
long time to diverge from its relatives? he most reliable approach is to
compare several genes from the same two species, as shown for rat and
human in Table 1–1. Two measures of rates of nucleotide substitution
are indicated in the table. Nonsynonymous changes refer to single nucleotide changes in the DNA sequence that alter the encoded amino acid
(ATC → TTC, which is I → F, for example). Synonymous changes refer
to those that do not alter the encoded amino acid (ATC → ATT, which is
I → I, for example). (As is apparent in the genetic code, see Tables 7 and
8 on page 966, individual amino acids are typically encoded by multiple
codons.)
A. Why are there such large diferences between the synonymous and nonsynonymous rates of nucleotide substitution?
B. Considering that the rates of synonymous changes are about the same
for all three genes, how is it possible for the histone H3 gene to resist so
efectively those nucleotide changes that alter the amino acid sequence?
C. In principle, a gene might be highly conserved because it exists in a “privileged” site in the genome that is subject to very low mutation rates. What
feature of the data in Table 1–1 argues against this possibility for the histone H3 gene?
TABLE 1–1 Rates of nucleotide substitutions in three genes from rat
and human (Problem 1–49).
Rates of change
Gene
Amino
acids
Nonsynonymous
Synonymous
Histone H3
135
0.0
4.5
Hemoglobin α
141
0.6
4.4
Interferon γ
136
3.1
5.5
Rates are expressed as nucleotide changes per site per 109 years. The
average rate of nonsynonymous changes for several dozen rat and human
genes is about 0.8.
1–50
Plant hemoglobins were found initially in legumes, where they function
in root nodules to lower the oxygen concentration so that the resident
bacteria can ix nitrogen. hese hemoglobins impart a characteristic pink
color to the root nodules. When these genes were irst discovered, it was
GENETIC INFORMATION IN EUKARYOTES
9
Figure 1–3 Phylogenetic tree for
hemoglobin genes from a variety of
species (Problem 1–50). The legumes
are shown in green. The lengths of lines
that connect the present-day species
represent the evolutionary distances that
separate them.
VERTEBRATES
Salamander
Cobra
Rabbit
Chicken
Whale
Cat
Human
Cow
Frog
Goldfish
Barley
Earthworm
Lotus
PLANTS
Alfalfa
Insect
Bean
Clam
Chlamydomonas
INVERTEBRATES
Nematode
PROTOZOA
Paramecium
so surprising to ind a gene typical of animal blood that it was hypothesized that the plant gene arose by horizontal transfer from some animal.
Many more hemoglobin genes have now been sequenced, and a phylogenetic tree based on some of these sequences is shown in Figure 1–3.
A. Does this tree support or refute the hypothesis that the plant hemoglobins arose by horizontal gene transfer?
B. Supposing that the plant hemoglobin genes were originally derived from
a parasitic nematode, for example, what would you expect the phylogenetic tree to look like?
MCAT STYLE
Passage 1 (Questions 1–51 to 1–53)
Molecules found in nature (termed “natural products”) are a rich source of compounds with biological activities that are useful as drugs. Marine organisms are a
particularly good source of natural products. Imagine that you are a marine biologist searching for new natural products for use as anticancer drugs. While diving
at a coral reef near Tahiti, you notice an unusual sponge that you have never seen
before. You collect a sample, take it back to your lab, make an extract, and ind that
it contains a compound that kills cancer cells, but not normal cells. You want to
learn more about the active compound and how it is produced. Based on previous studies, you suspect that it may come from bacteria or archaea that live in a
symbiotic relationship with the sponge, which is a eukaryote.
1–51
To deine the source of the compound, you initially want to determine
whether the sponge sample contains bacteria or archaea, and if so, how
many diferent kinds. What would be the most informative way to do
this?
A. Compare the sequences of the genes that encode ribosomal RNA for all
the cells that are part of the sponge sample.
B. Culture each of the microorganisms and then test their susceptibility to
antibiotics that kill bacteria or archaea.
10
Chapter 1: Cells and Genomes
C. Culture the microorganisms and classify them according to their nutritional requirements and biochemical pathways.
D. Examine the microorganisms under a microscope to see whether they
look like bacteria or like archaea.
1–52
You are eventually able to culture multiple kinds of bacteria from the
sponge. To your surprise, you discover that two very diferent bacteria
produce the same anticancer compound. You sequence their genomes
and ind that the vast majority of their genes are either unique to each
kind of bacterium or show signiicant sequence diferences. However,
several genes are nearly identical. You suspect that one or more of the
nearly identical genes are involved in producing the compound. How is
it that such divergent bacteria could have genes that are nearly identical?
A. he bacteria obtained the genes by horizontal transfer from the sponge.
B. he bacteria shared the genes by horizontal transfer of a plasmid.
C. he genes are from a common ancestor of the two kinds of bacteria.
D. he genes became nearly identical by convergent evolution.
1–53
You identify a gene that may play an essential role in the biosynthetic
pathway that makes the compound. What would be the fastest way to
gain clues to the function or biochemical activity of the protein encoded
by this gene?
A. Compare the gene’s sequence to all other gene sequences to see if it is
similar to known genes.
B. Express and purify the protein encoded by the gene and study its enzyme
functions in vitro.
C. Express the gene in E. coli and determine whether E. coli can now make
the compound.
D. Make mutations in the gene and determine how they afect synthesis of
the compound.
Chapter 2
11
CHAPTER
Cell Chemistry and
Bioenergetics
2
THE CHEMICAL COMPONENTS OF A CELL
TERMS TO LEARN
acid
base
buffer
chemical group
covalent bond
electrostatic attraction
hydrogen bond
hydronium ion (H3O+)
hydrophilic
hydrophobic
hydrophobic force
macromolecule
pH scale
proton (H+)
van der Waals attraction
DEFINITIONS
IN THIS CHAPTER
THE CHEMICAL COMPONENTS
OF A CELL
CATALYSIS AND THE USE OF
ENERGY BY CELLS
HOW CELLS OBTAIN ENERGY
FROM FOOD
Match each deinition below with its term from the list above.
2–1
Force exerted by the hydrogen-bonded network of water molecules that
brings two nonpolar surfaces together by excluding water between them.
2–2
Noncovalent bond in which an electropositive hydrogen atom is partially
shared by two electronegative atoms.
2–3
Substance that releases protons when dissolved in water, forming a
hydronium ion (H3O+).
2–4
Type of noncovalent bond that is formed at close range between nonpolar atoms.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
2–5
A 10–8 M solution of HCl has a pH of 8.
Strong acids bind protons strongly.
2–7
Most of the interactions between macromolecules could be mediated
just as well by covalent bonds as by noncovalent bonds.
THOUGHT PROBLEMS
2–8
2–9
he mass of a hydrogen atom—and thus of a proton—is almost exactly
1 dalton. If protons and neutrons have virtually identical masses, and
the mass of an electron is negligible, shouldn’t all elements have atomic
weights that are nearly integers? A perusal of the periodic table shows
that this simple expectation is not true. Chlorine, for example, has an
atomic weight of 35.5. How is it that elements can have atomic weights
that are not integers?
C, H, and O account for 95% of the elements in living organisms (Figure
2–1). hese atoms are present in the ratio C:2H:O, which is equivalent
to the general formula for carbohydrates (CH2O). Does this mean living
organisms are mostly sugar? Why or why not?
50
relative abundance (percent)
2–6
40
30
20
10
0
H
C
O
N Ca Na P others
& &
Mg K
elements
Figure 2–1 Abundance of elements in
living organisms (Problem 2–9).
12
Chapter 2: Cell Chemistry and Bioenergetics
Figure 2–2 Space-illing models of H2o
and H2S (Problem 2–11).
water (H2O)
hydrogen sulfide (H2S)
2–10
Order the following list of processes in terms of their energy content from
smallest to largest.
A. ATP hydrolysis in cells.
B. Average thermal motions.
C. C–C bond.
D. Complete oxidation of glucose.
E. Noncovalent bond in water.
2–11
Oxygen and sulfur have similar chemical properties because both elements have six electrons in their outermost electron shells. Indeed, both
oxygen and sulfur form molecules with two hydrogen atoms: water (H2O)
and hydrogen sulide (H2S) (Figure 2–2). Surprisingly, water is a liquid,
yet H2S is a gas, even though sulfur is much larger and heavier than oxygen. Propose an explanation for this striking diference.
2–12
What do you think the “p” in pH stands for?
2–13
Imagine that you dissolve some crystals of sodium chloride, potassium
acetate, and ammonium chloride in separate beakers of water. Predict
whether the pH values of the resulting solutions would be acidic, neutral,
or basic. Explain your reasoning.
2–14
In solution at pH = 1, the amino acid glycine (+H3NCH2COOH) has two
ionizable groups: the carboxylic acid group (–COOH) and the basic
amine group (–NH3+). Adding NaOH to this solution gives the titration
curve shown in Figure 2–3.
A. Write the expressions (HA H+ + A–) for dissociation of the carboxylic
acid (–COOH) and amine groups (–NH3+).
B. Recall that pK is the pH at which exactly half of the carboxylic acid or
amine groups are charged. Estimate the pK values for the carboxylate
and amine groups of glycine.
C. Indicate the predominant ionic species of glycine at each of the points
shown on the curve in Figure 2–3.
D. he isoelectric point of a solute is the pH at which it carries no net charge.
Estimate the isoelectric point for glycine from the curve in Figure 2–3.
2–15
If you want to order glycine from a chemical supplier, you have three
choices: glycine, glycine sodium salt, and glycine hydrochloride. Write
the structures of these three compounds.
2–16
From the pK values listed in Table 2–1, decide which amino acids were
used in the titration curves shown in Figure 2–4.
12
10
8
pH 6
4
2
2–17
During an all-out sprint, muscles metabolize glucose anaerobically,
producing a high concentration of lactic acid, which lowers the pH of
the blood and of the cytosol. he lower pH inside the cell decreases the
eiciency of certain glycolytic enzymes, which reduces the rate of ATP
production and contributes to the fatigue that sprinters experience well
0
0
0.5
1.0
1.5
2.0
NaOH added (equivalents)
Figure 2–3 Titration of a solution of
glycine (Problem 2–14). one equivalent of
oH– is the amount required to completely
neutralize one acidic group.
THE CHEMICAL COMPONENTS OF A CELL
(A)
13
(B)
14
Figure 2–4 Titration curves for two amino
acids (Problem 2–16).
14
12
12
10
10
8
8
pH
pH
6
6
4
4
2
2
0
0
1.0
2.0
OH– added (equivalents)
0
3.0
0
1.0
2.0
OH– added (equivalents)
3.0
TABLE 2–1 Values for the ionizable groups of several amino acids
(Problem 2–16).
pK Values
–CooH
–NH3+
Leucine
2.4
9.6
Proline
2.0
10.6
Glutamate
2.2
9.7
4.3 (carboxyl)
Histidine
1.8
9.2
6.0 (imidazole)
Cysteine
1.8
10.8
8.3 (sulfhydryl)
Arginine
1.8
9.0
12.5 (guanidino)
Lysine
2.2
9.2
10.8 (amino)
Amino acid
R Group
before their fuel reserves are exhausted. he main blood bufer against
pH changes is the bicarbonate/CO2 system.
pK1 =
2.3
pK2 =
3.8
CO2
CO2
H2CO3
(gas) (dissolved)
pK3 =
– 10.3
H+ + HCO3
H+ + CO32–
To improve their performance, would you advise sprinters to hold their
breath or to breathe rapidly for a minute immediately before the race?
Explain your answer.
2–18
Aspirin is a weak acid (Figure 2–5) that is taken up into the bloodstream
by difusion through cells lining the stomach and the small intestine.
Aspirin crosses the plasma membrane of a cell most efectively in its
uncharged form; in its charged form it cannot cross the hydrophobic
lipid bilayer of the membrane. he pH of the stomach is about 1.5 and
that of the lumen of the small intestine is about 6.0. Is the majority of the
aspirin absorbed in the stomach or in the intestine? Explain your reasoning.
O
O
C
H3 C
O
C
O
C
uncharged form
pK = 3.5
OH
H3 C
O
O
C
charged form
O–
+
H+
Figure 2–5 Aspirin (Problem 2–18).
14
Chapter 2: Cell Chemistry and Bioenergetics
2–19
What, if anything, is wrong with the following statement: “When NaCl is
dissolved in water, the water molecules closest to the ions will tend to orient themselves so that their oxygen atoms point toward the sodium ions
and away from the chloride ions.” Explain your answer.
O
–O
O
If noncovalent interactions are so weak in a water environment, how can
they possibly be important for holding molecules together in cells?
HO CH
2–21
he three molecules in Figure 2–6 contain the seven most common reactive groups in biology. Most molecules in the cell are built from these
functional groups. Indicate and name the functional groups in these
molecules.
–O
CH2
O
2–24
P
O
O–
1,3-bisphosphoglycerate
here are many diferent, chemically diverse ways in which small molecules can be linked to form polymers. For example, ethene (CH2=CH2)
is used commercially to make the plastic polymer polyethylene (…–CH2–
CH2–CH2–CH2–…). he individual subunits of the three major classes of
biological macromolecules, however, are all linked by similar reaction
mechanisms; namely, by condensation reactions that eliminate water.
Can you think of any beneits that this chemistry ofers and why it might
have been selected in evolution?
O–
O
C
C
O
CH3
pyruvate
CALCULATIONS
2–23
O
C
2–20
2–22
O–
P
SH
To gain a better feeling for atomic dimensions, assume that the page
on which this question is printed is made entirely of the polysaccharide cellulose (Figure 2–7). Cellulose is described by the formula
(C6H12O6)n, where n is a large number that varies from one molecule to
another. he atomic weights of carbon, hydrogen, and oxygen are 12, 1,
and 16, respectively, and this page weighs 5 grams.
A. How many carbon atoms are there in this page?
B. In paper made of pure cellulose, how many carbon atoms would be
stacked on top of each other to span the thickness of this page (the page
is 21 cm × 27.5 cm × 0.07 mm)? (Rather than solving three simultaneous equations for the carbon atoms in each dimension, you might try a
shortcut. Determine the linear density of carbon atoms by calculating the
number of carbon atoms on the edge of a cube with the same volume as
this page, and then adjust that number to the thickness of the page.)
C. Now consider the problem from a diferent angle. Assume that the page
is composed only of carbon atoms, which have a van der Waals radius
of 0.2 nm. How many carbon atoms stacked end to end at their van der
Waals contact distance would it take to span the thickness of the page?
D. Compare your answers from parts B and C and explain any diferences.
CH2
O
C
CH
NH3 +
O–
cysteine
Figure 2–6 Three molecules that illustrate
the seven most common functional
groups in biology (Problem 2–21).
1,3-Bisphosphoglycerate and pyruvate
are intermediates in glycolysis and
cysteine is an amino acid.
In the United States, the concentration of glucose in blood is commonly
reported in milligrams per deciliter (dL = 100 mL). Over the course of a
day in a normal individual the circulating levels of glucose vary around a
mean of about 90 mg/dL (Figure 2–8). What would this value be if it were
expressed as a molar concentration of glucose in blood, which is the way
it is typically reported in the rest of the world?
HO
HO
CH2
OH
O
O
HO
HO
OH
CH2
OH
O
O
CH2
OH
HO
O
O
Figure 2–7 Structure of the polysaccharide cellulose (Problem 2–23).
HO
HO
OH
CH2
OH
O
O
CH2
OH
HO
O
O
O
CH2
OH
O
HO
OH
THE CHEMICAL COMPONENTS OF A CELL
CH2 OH
H
C
HO
C
H
O
OH
H
OH
C
H
C
C
H
OH
Figure 2–8 Circulating blood glucose
(Problem 2–24). (A) Structure of glucose.
(B) Typical variation in blood glucose over
the course of a day.
(B) GLUCOSE LEVELS IN BLOOD
blood glucose (mg/dL)
(A) STRUCTURE OF GLUCOSE
15
breakfast
150
dinner
100
50
lunch
0
3
6
9
12
15
18
21
24
3
hour of day
2–25
Imagine that you have a beaker of pure water at neutral pH (pH 7.0).
A. What is the concentration of H3O+ ions and how were they formed?
B. What is the molarity of pure water? (Hint: 1 liter of water weighs 1 kg.)
C. What is the ratio of H3O+ ions to H2O molecules?
2–26
By a convenient coincidence the ion product of water, Kw = [H+][OH–], is
a nice round number: 1.0 × 10–14 M2.
A. Why is a solution at pH 7.0 said to be neutral?
B. What is the H+ concentration and pH of a 1 mM solution of NaOH?
C. If the pH of a solution is 5.0, what is the concentration of OH– ions?
he Henderson–Hasselbalch equation
2–27
pH = pK + log
[A–]
[HA]
is a useful transformation of the equation for dissociation of a weak acid,
HA:
[H+][A–]
K =
[HA]
A. It is instructive to use the Henderson–Hasselbalch equation to determine
the extent of dissociation of an acid at pH values above and below the pK.
For the pH values listed in Table 2–2, ill in the values for log [A–]/[HA]
and [A–]/[HA], and indicate the percentage of the acid that has dissociated.
TABLE 2–2 Dissociation of a weak acid at pH values above and below the
pK (Problem 2–27).
pH
pK +4
pK +3
pK +2
pK +1
pK
pK –1
pK –2
pK –3
pK –4
log
[A–]
[HA]
[A–]
[HA]
% Dissociation
16
Chapter 2: Cell Chemistry and Bioenergetics
+4
Figure 2–9 Graph for plotting values from
Table 2–2 (Problem 2–27).
+3
+2
+1
pH pK
–1
–2
–3
–4
0
10
20
30
all HA
40
50
60
70
80
90
100
—
all A
percent dissociated
B. Using the graph in Figure 2–9, sketch the relationship between pH of the
solution and the fractional dissociation of a weak acid. Will the shape of
this curve be the same for all weak acids?
2–28
Cells maintain their cytosolic pH in a narrow range around pH 7.0 by
using a variety of weak acids to bufer against changes in pH. his is
essential because a large number of processes in cells generate or consume H+ ions. Weak acids resist changes in pH—the deinition of a
bufer—most efectively within about one pH unit on either side of their
pK values. Ionization of phosphoric acid provides an important bufering
system in cells. Phosphoric acid has three ionizable protons, each with a
unique pK.
H3PO4
pK = 2.1
H+ + H2PO4–
pK = 6.9
H+ + HPO42–
pK = 12.4
H+ + PO43–
A. Using the values derived in Problem 2–27, estimate how much of each
of the four forms of phosphate (H3PO4, H2PO4–, HPO42–, and PO43–), as
a percentage of the total, are present in the cytosol of cells at pH 7. (No
calculators permitted.)
B. What is the ratio of [HPO42–] to [H2PO4–] ([A–]/[HA]) in the cytosol at pH
7? If the cytosol is 1 mM phosphate (sum of all forms), what are the concentrations of H2PO4– and HPO42– in the cytosol? (Calculators permitted.)
2–29
Inside cells, the two most important bufer systems are provided by phosphate and proteins. he quantitative aspects of a bufer system pertain
to both the efective bufering range (how near the pH is to the pK for
the bufer) and the overall concentration of the bufering species (which
determines the number of protons that can be handled). As discussed in
Problem 2–28, H2PO4– HPO42– has a pK of 6.9 with an overall intracellular phosphate concentration of about 1 mM. In red blood cells, the
concentration of globin chains (molecular weight = 15,000) is about 100
mg/mL and each has 10 histidines, with pK values between 6.5 and 7.0.
Which of these two bufering systems do you think is quantitatively the
more important in red blood cells, and why do you think so?
THE CHEMICAL COMPONENTS OF A CELL
2–30
he most important bufer in the bloodstream is the bicarbonate/CO2
system. It is much more important than might be expected from its pK
because it is an open system in which the CO2 is maintained at a relatively constant value by exchange with the atmosphere. (By contrast, the
bufering systems described in Problem 2–29 are closed systems with no
exchange.) he equilibria involved in the bicarbonate/CO2 bufering system are
pK1 =
pK2 =
pK3 =
2.3
3.8
10.3 +
CO2(gas)
CO2(dis)
H2CO3
H+ + HCO3–
H + CO32–
H+ + CO32–) is so high that it never comes into play in
pK3 (HCO3–
biological systems. pK2 (H2CO3 H+ + HCO3–) seems much too low to
be useful, but it is inluenced by the dissolved CO2 [CO2(dis)], which is
directly proportional to the partial pressure of CO2 in the gas phase. he
dissolved CO2 in turn is kept in equilibrium with H2CO3 by the enzyme
carbonic anhydrase.
K1 =
[H2CO3]
= 5 × 10–3, or pK1 = 2.3
[CO2(dis)]
he equilibrium for hydration of CO2(dis) can be combined with the
equilibrium for dissociation of H2CO3 to give a pKʹ for CO2(dis) H+ +
HCO3–
[H+][HCO3–]
K′ =
= K1 × K2
[CO2(dis)]
pK′ = pK1 + pK2 = 2.3 + 3.8 = 6.1
Even the pK ʹ of 6.1 seems too low to maintain the blood pH around 7.4,
yet this open system is very efective, as can be illustrated by a few calculations. he total concentration of carbonate in its various forms, but
almost entirely CO2(dis) and HCO3–, is about 25 mM.
A. Using the Henderson–Hasselbalch equation, calculate the ratio of
HCO3– to CO2(dis) at pH 7.4. What are the concentrations of HCO3– and
CO2(dis)?
B. What would the pH be if 5 mM H+ were added under conditions where
CO2 was not allowed to leave the system; that is, if the concentration of
CO2(dis) was not maintained at a constant value?
C. What would the pH be if 5 mM H+ were added under conditions where
CO2 was permitted to leave the system; that is, if the concentration of
CO2(dis) was maintained at a constant value?
2–31
he proteins in a mammalian cell account for 18% of its net weight. If the
density of a typical mammalian cell is about 1.1 g/mL and the volume of
the cell is 4 × 10–9 mL, what is the concentration of protein in mg/mL?
DATA HANDLING
2–32
he ionizable groups in amino acids can inluence one another, as shown
by the pK values for the carboxyl and amino groups of alanine and various oligomers of alanine (Figure 2–10). Suggest an explanation for why
the pK of the carboxyl group increases with oligomer length, while that of
the amino group decreases.
2–33
A histidine side chain is known to have an important role in the catalytic
mechanism of an enzyme; however, it is not clear whether histidine is
required in its protonated (charged) or unprotonated (uncharged) state.
To answer this question you measure enzyme activity over a range of pH,
with the results shown in Figure 2–11. Which form of histidine is required
for enzyme activity?
17
Chapter 2: Cell Chemistry and Bioenergetics
CHEMICAL FORMULAS
pK VALUES
CH3
Ala
+
H3 N
COO–
CH
CH3
Ala2 +H3 N
CH
COO–
+
2.34
9.69
H3 N
CH3
C
NH
CH
COO–
3.12
8.30
O
CH3
Ala3 +H3 N
CH
CH3
C
NH
CH
O
CH
C
NH
CH3
C
CH
COO–
3.39
8.03
O
CH3
Ala4 +H3 N
CH3
NH
O
CH
CH3
CH3
C
NH
O
CH
C
NH
CH
COO–
3.42
7.94
O
Figure 2–10 pK values for the carboxyl and amino groups in oligomers of alanine
(Problem 2–32).
120
activity (% of maximum)
18
100
80
60
40
20
0
4
5
6
7
pH
8
9
10
Figure 2–11 Enzyme activity as a function
of pH (Problem 2–33).
MEDICAL LINKS
2–34
he drug thalidomide was once prescribed as a sedative to help with
nausea during the early stages of pregnancy. One of its optical isomers,
(R)-thalidomide (Figure 2–12), is the active agent responsible for its
sedative efects. It was synthesized, however, as a mixture of both optical isomers—a not uncommon practice that usually causes no problems.
Unfortunately, the other optical isomer is a teratogen that led to a horriic
series of birth defects characterized by malformed or absent limbs. On
the structural formula in Figure 2–12A, identify the carbon that is responsible for its optical activity (its chiral center) and sketch the structure of
the teratogenic form of thalidomide.
2–35
he molecular weight of ethanol (CH3CH2OH) is 46 and its density is
0.789 g/cm3.
A. What is the molarity of ethanol in beer that is 5% ethanol by volume?
[Alcohol content of beer varies from about 4% (lite beer) to 8% (stout
beer).]
B. he legal limit for a driver’s blood alcohol content varies, but 80 mg of
ethanol per 100 mL of blood (usually referred to as a blood alcohol level
of 0.08) is typical. What is the molarity of ethanol in a person at this legal
limit?
C. How many 12-oz (355-mL) bottles of 5% beer could a 70-kg person drink
and remain under the legal limit? A 70-kg person contains about 40 liters
of water. Ignore the metabolism of ethanol, and assume that the water
content of the person remains constant.
D. Ethanol is metabolized at a constant rate of about 120 mg per hour per
kg body weight, regardless of its concentration. If a 70-kg person were at
twice the legal limit (160 mg/100 mL), how long would it take for their
blood alcohol level to fall below the legal limit?
(A) THALIDOMIDE CHEMICAL FORMULA
(B) THALIDOMIDE SPACE-FILLING MODEL
O
N
O
O
N
H
O
Figure 2–12 The structure of the sedative
(R)-thalidomide (Problem 2–34).
CATALYSIS AND THE USE OF ENERGY BY CELLS
19
CATALYSIS AND THE USE OF ENERGY BY CELLS
TERMS TO LEARN
acetyl CoA
activated carrier
activation energy
ADP
aerobic respiration
ATP
catalyst
coupled reaction
diffusion
entropy
enzyme
equilibrium
equilibrium constant (K)
free energy (G)
free-energy change (∆G)
metabolism
NAD+/NADH
NADP+/NADPH
oxidation
reduction
standard free-energy
change (∆G°)
substrate
DEFINITIONS
Match each deinition below with its term from the list above.
2–36
Extra energy that must be possessed by atoms or molecules in addition
to their ground-state energy in order to undergo a particular chemical
reaction.
2–37
Free-energy change of two reacting molecules at standard temperature
and pressure when all components are present at a concentration of 1
mole per liter.
2–38
Loss of electrons from an atom, as occurs during the addition of oxygen
to a molecule or when a hydrogen is removed.
2–39
Molecule on which an enzyme acts.
2–40
Net drift of molecules in the direction of lower concentration due to random thermal movement.
2–41
Protein that catalyzes a speciic chemical reaction.
2–42
Linked pair of chemical reactions in which the free energy released by
one of the reactions serves to drive the other.
2–43
State at which there is no net change in a system. In a chemical reaction,
this state is reached when the forward and reverse rates are equal.
2–44
he energy that can be extracted from a system to drive reactions. Takes
into account changes in both energy and entropy.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
2–45
Animals and plants use oxidation to extract energy from food molecules.
2–46
If an oxidation occurs in a reaction, it must be accompanied by a reduction.
2–47
Linking the energetically unfavorable reaction A → B to a second, favorable reaction B → C will shift the equilibrium constant for the irst reaction.
THOUGHT PROBLEMS
2–48
he organic chemistry of living cells is said to be special for two reasons: it
occurs in an aqueous environment and it accomplishes some very complex reactions. But do you suppose it’s really all that much diferent from
the organic chemistry carried out in the top laboratories in the world?
Why or why not?
20
Chapter 2: Cell Chemistry and Bioenergetics
2–49
Distinguish between catabolic and anabolic pathways of metabolism,
and indicate in a general way how such pathways are linked to one
another in cells.
2–50
he second law of thermodynamics states that systems will change spontaneously toward arrangements with greater entropy (disorder). Living
systems are so intricately ordered, however, it seems they must surely
violate the second law. Explain briely—and in a way your parents could
understand—how life is fully compatible with the laws of thermodynamics.
2–51
In the reaction 2 Na + Cl2 → 2 Na+ + 2 Cl–, what is being oxidized and
what is being reduced? How can you tell?
2–52
If a cell in mitosis is cooled to 0°C, the microtubules in the spindle depolymerize into tubulin subunits. he same is true for microtubules made
from pure tubulin in a test tube; they assemble readily at 37°C, but disassemble at low temperature. In fact, many protein assemblies that are held
together by noncovalent bonds show the same behavior: they disassemble when cooled. his behavior is governed by the basic thermodynamic
equation
∆G = ∆H – T∆S
where ∆H is the change in enthalpy (chemical-bond energy), ∆S is the
change in entropy (disorder of the system), and T is the absolute temperature.
A. he change in free energy (∆G) must be negative for the reaction (tubulin
subunits → microtubules) to proceed at high temperature. At low temperature, ∆G must be positive to permit disassembly; that is, to favor the
reverse reaction. Decide what the signs (positive or negative) of ∆H and
∆S must be, and show how your choices account for polymerization of
tubulin at high temperature and its depolymerization at low temperature. (Assume that the ∆H and ∆S values themselves do not change with
temperature.)
B. Polymerization of tubulin subunits into microtubules at body temperature clearly occurs with an increase in the orderliness of the subunits
(Figure 2–13). Yet tubulin polymerization occurs with an increase in
entropy (decrease in order). How can that be?
POLYMERIZATION
2–53
Discuss the statement: “he criterion for whether a reaction proceeds
spontaneously is ∆G not ∆G°, because ∆G takes into account the concentrations of the substrates and products.”
2–54
At a particular concentration of substrates and products the reaction
below has a negative ∆G.
A + B → C + D ∆G = –20 kJ/mole
At the same concentrations, what is ∆G for the reverse reaction?
C+D→A+B
Figure 2–13 Polymerization of tubulin
subunits into a microtubule (Problem
2–52). The fates of one subunit (green)
and its associated water molecules (blue
spheres) are shown. Additional tubulin
subunits in the microtubule are lightly
shaded; their associated water molecules
are not shown.
CATALYSIS AND THE USE OF ENERGY BY CELLS
21
2–55
he values for ∆G° and for ∆G in cells have been determined for many
diferent metabolic reactions. What information do these values provide
about the rates of these reactions?
2–56
hermodynamically, it is perfectly valid to consider the cellular phosphorylation of glucose to produce glucose 6-phosphate (G6P) as the sum
of two reactions.
(1)
glucose + Pi → G6P + H2O
∆G° = 13.8 kJ/mole
(2)
ATP + H2O → ADP + Pi
∆G° = –30.5 kJ/mole
NET: glucose + ATP → G6P + ADP
But biologically it makes no sense at all. Hydrolysis of ATP (reaction 2)
in one part of the cell can have no efect on phosphorylation of glucose
(reaction 1) elsewhere in the cell, given that [ATP], [ADP], and [Pi] are
maintained within narrow limits. How does the cell manage to link these
two reactions to achieve the phosphorylation of glucose?
2–57
Each phosphoanhydride bond between the phosphate groups in ATP is a
high-energy linkage with a ∆G° value of –30.5 kJ/mole. Hydrolysis of this
bond in cells normally liberates usable energy in the range of 45 to 55 kJ/
mole. Why do you think a range of values for released energy is given for
∆G, rather than a precise number, as for ∆G°?
2–58
Consider the efects of two enzymes. Enzyme A catalyzes the reaction
ATP + GDP
ADP + GTP
whereas enzyme B catalyzes the reaction
NADH + NADP+
NAD+ + NADPH
Discuss whether the enzymes would be beneicial or detrimental to cells.
2–59
Match the activated carriers below with the group carried in high-energy
linkage.
A. Acetyl CoA
1. acetyl group
B. S-Adenosylmethionine
2. carboxyl group
C. ATP
3. electrons and hydrogens
D. Carboxylated biotin
4. glucose
E. NADH, NADPH, FADH2
5. methyl group
F. Uridine diphosphate glucose
6. phosphate
2–60
Which of the following reactions will occur only if coupled to a second,
energetically favorable reaction?
A. glucose + O2 → CO2 + H2O
B. CO2 + H2O → glucose + O2
C. nucleoside triphosphate + DNAn → DNAn+1 + 2 Pi
D. nucleosides → nucleoside triphosphates
E. ADP + Pi → ATP
CALCULATIONS
2–61
If an uncatalyzed reaction occurred at the rate of 1 event per century, and
if an enzyme speeded up the rate by a factor of 1014, how many seconds
would it take the enzyme to catalyze one event?
2–62
“Difusion” sounds slow—and over everyday distances it is—but on the
scale of a cell it is very fast. he average instantaneous velocity of a particle
in solution—that is, the velocity between its very frequent collisions—is
v = (kT/m)½
22
Chapter 2: Cell Chemistry and Bioenergetics
where k = 1.38 × 10–16 g cm2/K sec2, T = temperature in K (37°C is 310 K),
and m = mass in g/molecule.
Calculate the instantaneous velocity of a water molecule (molecular
mass = 18 daltons), a glucose molecule (molecular mass = 180 daltons),
and a myoglobin molecule (molecular mass = 15,000 daltons) at 37°C.
Just for fun, convert these numbers into kilometers/hour. Before you do
any calculations, you might try to guess whether the molecules are moving at a slow crawl (<1 km/hr), an easy walk (5 km/hr), or a record-setting
sprint (40 km/hr).
2–63
he instantaneous velocity tells you little about the time it takes for a
molecule to move cellular distances because its trajectory is constantly
altered by collisions with other molecules in solution (Figure 2–14). he
average time it takes for a molecule to travel x cm by difusion in three
dimensions is
start
t = x2/6D
where t is the time in seconds and D is the difusion coeicient, which
is a constant that depends on the size and shape of the particle. Glucose
and myoglobin, for example, have difusion coeicients of about 5 × 10–6
cm2/sec and 5 × 10–7 cm2/sec, respectively. Calculate the average time
it would take for glucose and myoglobin to difuse a distance of 20 μm,
which is approximately the width of a mammalian cell.
2–64
Phosphoglucose isomerase catalyzes the interconversion of glucose
6-phosphate (G6P) and fructose 6-phosphate (F6P):
G6P
F6P
he ∆G for this reaction is given by the equation
∆G = ∆G° + 2.3 RT log
[F6P]
[G6P]
where R = 8.3 × 10–3 kJ/K mole and T = 310 K. A useful number to remember is that 2.3 RT = 5.9 kJ/mole at 37°C, which is body temperature.
A. At equilibrium, [F6P]/[G6P] is equal to the equilibrium constant (K) for
the reaction. Rewrite the above equation for the reaction at equilibrium.
B. At equilibrium, the ratio of [F6P] to [G6P] is observed to be 0.5. At this
equilibrium ratio, what are the values of ∆G and ∆G°?
C. Inside a cell, the value of ∆G for this reaction is –2.5 kJ/mole. What is the
ratio of [F6P] to [G6P]? What is ∆G°?
2–65
A 70-kg adult human (154 lb) could meet his or her entire energy needs
for one day by eating 3 moles of glucose (540 g). (We don’t recommend
this.) Each molecule of glucose generates 30 molecules of ATP when it is
oxidized to CO2. he concentration of ATP is maintained in cells at about
2 mM, and a 70-kg adult has about 25 L of intracellular luid. Given that
the ATP concentration remains constant in cells, calculate how many
times per day, on average, each ATP molecule in the body is hydrolyzed
and resynthesized.
DATA HANDLING
2–66
he polymerization of subunits into a pentameric ring is shown in Figure 2–15. he equilibrium constants for association of a subunit at each
step in the assembly of the tetramer (that is, K1, K2, and K3) are approximately equal at 106 M–1. he equilibrium constant for association of the
inal subunit in the ring (K4), however, is >1012 M–1. Why is association
of the inal subunit so much more highly favored than association of the
initial subunits? Why do you suppose the equilibrium constant for the
finish
Figure 2–14 A two-dimensional,
simulated random walk of a molecule in
solution (Problem 2–63).
CATALYSIS AND THE USE OF ENERGY BY CELLS
+
K3
+
K2
+
+
K1
23
K4
Figure 2–15 Polymerization of subunits into a pentameric ring (Problem 2–66).
association of the inal subunit is approximately the square of the equilibrium constants for the earlier steps?
2–67
Red blood cells obtain energy in the form of ATP by converting glucose to
pyruvate via the glycolytic pathway (Figure 2–16). he values of ∆G° and
∆G (in kJ/mole) have been calculated for each of the steps in glycolysis in
red blood cells that are actively metabolizing glucose, as summarized in
Table 2–3. he ∆G° values are based on the known equilibrium constants
for the reactions; the ∆G values are calculated from the ∆G° values and
actual measurements of concentrations of the intermediates in red blood
cells.
he low of metabolites through a metabolic pathway can occur only
when the ∆G value for each step is negative. his is a true statement.
Despite this assertion, three reactions in red blood cell glycolysis have
slightly positive ∆G values. What do you suppose is the explanation for
the results in the table?
GLUCOSE
ATP
ADP + H+
G6P
F6P
ATP
ADP + H+
TABLE 2–3 The reactions of glycolysis in red blood cells and their associated
∆G° and ∆G values in kJ/mole (Problem 2–67).
Step
1
2
3
Reaction
∆G°
∆G
GLC + ATP → G6P + ADP + H+
–16.7
–33.4
+1.7
–2.5
–14.2
–22.2
G6P → F6P
F6P + ATP → F1,6BP + ADP + H+
F1,6BP → DHAP + G3P
+23.8
–1.3
5
DHAP → G3P
+7.5
+2.5
+6.3
–1.7
–18.8
+1.3
7
G3P + Pi + NAD+ → 1,3BPG + NADH + H+
1,3BPG + ADP → 3PG + ATP
DHAP
G3P
Pi + NAD+
NADH + H+
1,3BPG
ADP
4
6
F1,6BP
ATP
8
3PG → 2PG
+4.6
+0.8
9
2PG → PEP + H2O
+1.7
–3.3
10
PEP + ADP + H+ → PYR + ATP
–31.4
–16.7
3PG
2PG
H2O
PEP
GLC = glucose; G6P = glucose 6-phosphate; F6P = fructose 6-phosphate;
F1,6BP = fructose 1,6-bisphosphate; DHAP = dihydroxyacetone phosphate;
G3P = glyceraldehyde 3-phosphate; 1,3BPG = 1,3-bisphosphoglycerate;
3PG = 3-phosphoglycerate; 2PG = 2-phosphoglycerate;
PEP = phosphoenolpyruvate; PYR = pyruvate.
ADP + H+
ATP
PYRUVATE
Figure 2–16 The glycolytic pathway
(Problem 2–67). See Table 2–3 for the key
to abbreviations.
24
Chapter 2: Cell Chemistry and Bioenergetics
HOW CELLS OBTAIN ENERGY FROM FOOD
TERMS TO LEARN
citric acid cycle
electron-transport chain
FAD/FADH2
fat
fermentation
glycogen
glycolysis
GTP
nitrogen fixation
oxidative phosphorylation
starch
DEFINITIONS
Match each deinition below with its term from the list above.
2–68
Central metabolic pathway found in aerobic organisms, which oxidizes
acetyl groups derived from food molecules to CO2 and H2O. In eukaryotic cells, it occurs in the mitochondria.
2–69
Energy-storage lipids in cells that are composed of triacylglycerols (triglycerides), which are fatty acids esteriied with glycerol.
2–70
Polysaccharide composed exclusively of glucose units used to store
energy in animal cells. Granules of it are especially abundant in liver and
muscle cells.
2–71
Process in bacteria and mitochondria in which ATP formation is driven
by the transfer of electrons from food molecules to molecular oxygen.
2–72
Series of electron carrier molecules along which electrons move from a
higher to a lower energy level to a inal acceptor molecule, with the associated production of ATP.
2–73
Ubiquitous metabolic pathway in the cytosol in which sugars are partially metabolized to produce ATP.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
2–74
Because glycolysis is only a prelude to the oxidation of glucose in mitochondria, which yields 15-fold more ATP, glycolysis is not really important for human cells.
2–75
he reactions of the citric acid cycle do not directly require the presence
of oxygen.
THOUGHT PROBLEMS
2–76
Match the polymeric molecules in food with the monomeric subunits
into which they are digested before they can be oxidized to produce
energy.
A. fats
1. amino acids
B. polysaccharides
2. fatty acids
C. proteins
3. glycerol
4. sugars
2–77
From a chemical perspective, the glycolytic pathway (see Figure 2–16)
can be thought of as occurring in two stages. he irst stage from glucose
to glyceraldehyde 3-phosphate (G3P) prepares glucose so that its cleavage yields G3P and an equivalent three-carbon fragment, which is then
converted into G3P. he second stage—from G3P to pyruvate—harvests
energy in the form of ATP and NADH.
A. Write the balanced equation for the irst stage of glycolysis (glucose →
G3P).
HOW CELLS OBTAIN ENERGY FROM FOOD
25
B. Write the balanced equation for the second stage of glycolysis (G3P →
pyruvate).
C. Write the balanced equation for the overall pathway (glucose → pyruvate).
2–78
2–79
2–80
GLUCOSE
ATP
ADP + H+
G6P
At irst glance, fermentation of pyruvate to lactate appears to be an
optional add-on reaction to glycolysis (Figure 2–17). After all, couldn’t
cells growing in the absence of oxygen simply discard pyruvate as a waste
product? In the absence of fermentation, which products derived from
glycolysis would accumulate in cells under anaerobic conditions? Could
the metabolism of glucose via the glycolytic pathway continue in the
absence of oxygen in cells that cannot carry out fermentation? Why or
why not?
In the absence of oxygen, cells consume glucose at a high, steady rate.
When oxygen is added, glucose consumption drops precipitously and is
then maintained at the lower rate. Why is glucose consumed at a high
rate in the absence of oxygen and at a low rate in its presence?
F6P
ATP
ADP + H+
F1,6BP
DHAP
G3P
Pi + NAD+
NADH + H+
Arsenate (AsO43–) is chemically very similar to phosphate (PO43–) and is
1,3BPG
used as an alternative substrate by many phosphate-requiring enzymes.
In contrast to phosphate, however, the anhydride bond between arsenate
and a carboxylic acid group is very quickly hydrolyzed in water. Knowing
this, suggest why arsenate is a compound of choice for murderers, but
not for cells. Formulate your explanation in terms of the step in glycolysis
at which 1,3-bisphosphoglycerate is converted to 3-phosphoglycerate,
generating ATP (Figure 2–18).
2–81
2–82
he liver provides glucose to the rest of the body between meals. It does
so by breaking down glycogen, forming glucose 6-phosphate in the
penultimate step. Glucose 6-phosphate is converted to glucose by splitting of the phosphate (∆G° = –13.8 kJ/mole). Why do you suppose the
liver removes the phosphate by hydrolysis, rather than reversing the
reaction by which glucose 6-phosphate is formed from glucose (glucose
+ ATP → G6P + ADP, ∆G° = –16.7 kJ/mole)? By reversing this reaction, the
liver could generate both glucose and ATP.
What, if anything, is wrong with the following statement: “he oxygen
consumed during the oxidation of glucose in animal cells is returned
as CO2 to the atmosphere.” How might you support your answer experimentally?
CALCULATIONS
2–83
If a cell hydrolyzes and replaces 109 ATP molecules per minute, how long
will it take for a cell to consume its own volume (1000 μm3) of oxygen?
Roughly 90% of the ATP in the cell is regenerated by oxidative phosphorylation. Assume that 5 molecules of ATP are regenerated by each
1,3-bisphosphoglycerate
ADP
ATP
3PG
2PG
H2O
PEP
ADP + H+
NADH + H+
ATP
NAD+
PYRUVATE
NAD+
CoA
NADH
NAD+
CO2
AcCoA
LACTATE
citric acid
cycle
fermentation
pathway
(aerobic)
(anaerobic)
Figure 2–17 Fermentation of pyruvate
to lactate (Problem 2–78). AcCoA, acetyl
CoA. See Table 2-3 for the key to the rest
of the abbreviations.
3-phosphoglycerate
O
O
O
O–
C
ADP
CH
ATP
HO
CH2
P
O–
CH
CH2
O
–O
O–
O
O–
C
HO
P
O
O
–O
P
O–
O
Figure 2–18 Conversion of
1,3-bisphosphoglycerate and ADP to
3-phosphoglycerate and ATP (Problem 2–80).
26
Chapter 2: Cell Chemistry and Bioenergetics
molecule of oxygen (O2) that is converted to water. Recall that a mole of a
gas occupies 22.4 L.
2–84
Assuming that there are 5 × 1013 cells in the human body and that ATP is
turning over at a rate of 109 ATP molecules per minute in each cell, how
many watts is the human body consuming? (A watt is a J per second.)
Assume that hydrolysis of ATP yields 50 kJ/mole.
2–85
Does a Snickers™ candy bar (65 g, 1360 kJ) provide enough energy to
climb from Zermatt (elevation 1660 m) to the top of the Matterhorn (4478
m, Figure 2–19), or might you need to stop at HÖrnli Hut (3260 m) to eat
another one? Imagine that you and your gear have a mass of 75 kg, and
that all of your work is done against gravity (that is, you’re just climbing
straight up).
work (J) = mass (kg) × g (m/sec2) × height gained (m)
where g is acceleration due to gravity (9.8 m/sec2). One joule is 1 kg m2/
sec2.
What assumptions made here will greatly underestimate how much
candy you need?
2–86
Muscles contain creatine phosphate (CP) as an energy bufer to maintain
the levels of ATP in the initial stages of exercise. Creatine phosphate can
transfer its phosphate to ADP to generate creatine (C) and ATP, with a
∆G° of –13.8 kJ/mole.
CP + ADP → C + ATP
∆G° = –13.8 kJ/mole
A. In a resting muscle, [ATP] = 4 mM, [ADP] = 0.013 mM, [CP] = 25 mM, and
[C] = 13 mM. What is the ∆G for this reaction in resting muscle? Does this
value make sense to you? Why or why not?
B. Consider an initial stage in vigorous exercise, when 25% of the ATP has
been converted to ADP. Assuming that no other concentrations have
changed, what is the ∆G for the reaction at this stage in exercising muscle? Does this value make sense?
C. If the ATP in muscle could be completely hydrolyzed (in reality it never
is), it would power an all-out sprint for about 1 second. If creatine phosphate could be completely hydrolyzed to regenerate ATP, how long could
a sprint be powered? Where do you suppose the energy comes from to
allow a runner to inish a 200-meter sprint?
DATA HANDLING
2–87
In 1904, Franz Knoop performed what was probably the irst successful
labeling experiment to study metabolic pathways. He fed many diferent fatty acids labeled with a terminal benzene ring to dogs and analyzed
their urine for excreted benzene derivatives. Whenever the fatty acid had
an even number of carbon atoms, phenylacetate was excreted (Figure
2–20A). Whenever the fatty acid had an odd number of carbon atoms,
benzoate was excreted (Figure 2–20B).
From these experiments Knoop deduced that oxidation of fatty acids
to CO2 and H2O involved the removal of two-carbon fragments from the
carboxylic acid end of the chain. Can you explain the reasoning that led
him to conclude that two-carbon fragments, as opposed to any other
number, were removed, and that degradation was from the carboxylic
acid end, as opposed to the other end?
2–88
In 1937, Hans Krebs deduced the operation of the citric acid cycle (Figure 2–21) from careful observations on the oxidation of carbon compounds in minced preparations of pigeon light muscle. (Pigeon breast
is a rich source of mitochondria, but the function of mitochondria was
Figure 2–19 The Matterhorn (Problem 2–85).
HOW CELLS OBTAIN ENERGY FROM FOOD
27
Figure 2–20 The original labeling
experiment to analyze fatty acid oxidation
(Problem 2–87). (A) Fed and excreted
derivatives of an even-number fatty acid
chain. (B) Fed and excreted derivatives of
an odd-number fatty acid chain.
O
(A)
fed
compound
CH2 CH2 CH2 CH2 CH2 CH2 CH2 C
excreted
compound
CH2
O–
eight-carbon chain
O
C
O–
phenylacetate
O
(B)
fed
compound
CH2 CH2 CH2 CH2 CH2 CH2
excreted
compound
C
C
O–
seven-carbon chain
O
O–
benzoate
unknown at the time.) In one set of experiments, Krebs found that addition of a small amount of citrate resulted in a much larger increase in
the consumption of oxygen than could be accounted for by the oxidation
of citrate (Table 2–4). his surprising observation ultimately led to the
description of the citric acid cycle.
A. If citrate were an intermediate in a linear pathway of oxidation, would
you expect that addition of a small amount would lead to a large increase
in oxygen consumption? Why or why not?
PYR
AcCoA
CoA
NADH
OAA
CIT
NAD+
H2O
MAL
ICIT
H2O
NAD+
NADH + CO2
CoA
FUM
αKG
CoA
NAD
FADH2
SUC
Figure 2–21 The citric acid cycle (Problem
2–88). PyR = pyruvate, AcCoA = acetyl
coenzyme A, CIT = citrate, ICIT =
isocitrate, αkG = α-ketoglutarate, ScCoA
= succinyl coenzyme A, SUC = succinate,
FUM = fumarate, MAl = malate, and
oAA = oxaloacetate.
+
ScCoA
NADH + CO2
FAD
GTP
GDP + Pi
TABLE 2–4 Respiration in minced pigeon breast in the presence and absence of
citrate (Problem 2–88).
oxygen consumption (mmol)
Time (minutes)
No citrate
3 mmol citrate
Difference
30
29
31
2
60
47
68
21
90
51
87
36
150
53
93
40
Chapter 2: Cell Chemistry and Bioenergetics
28
B. How does the operation of the citric acid cycle explain the high level of
oxygen consumption after addition of a small amount of citrate?
C. Toward the end of the paper, Krebs states, “While the citric acid cycle
thus seems to occur generally in animal tissues, it does not exist in yeast
or in E. coli, for yeast and E. coli do not oxidize citric acid at an appreciable rate.” Why do you suppose Krebs got this point wrong?
2–89
Pathways for synthesis of amino acids in microorganisms were worked
out in part by cross-feeding experiments among mutant organisms
that were defective for individual steps in the pathway. Results of crossfeeding experiments for three mutants defective in the tryptophan pathway—TrpB–, TrpD–, and TrpE–—are shown in Figure 2–22A. he mutants
were streaked on a Petri dish and allowed to grow briely in the presence
of a very small amount of tryptophan, producing three pale streaks. As
shown, heavier growth was observed at points where some streaks were
close to other streaks. hese spots of heavier growth indicate that one
mutant can cross-feed (supply an intermediate) to the other one.
A. From the pattern of cross-feeding shown in Figure 2–22A, deduce the
order of the steps controlled by the products of the TrpB, TrpD, and TrpE
genes. Explain your reasoning.
B. If accumulated intermediates at the block are responsible for the crossfeeding phenomenon, it should be possible to grow individual mutants
on some intermediates. he three mutants were tested for growth on
tryptophan and intermediates in the pathway (Figure 2–22B), with the
results shown in Table 2–5. Use this information to arrange the defective
genes relative to the tryptophan pathway.
TABLE 2–5 Growth of mutants on intermediates in the pathway for tryptophan
biosynthesis (Problem 2–89).
Strain
Growth on minimal medium supplemented with
None
Chorismate
Anthranilate
Indole
Tryptophan
Wild type
+
+
+
+
+
TrpB–
–
–
–
–
+
TrpD–
–
–
–
+
+
TrpE–
–
–
+
+
+
MCAT STYLE
Passage 1 (Questions 2–90 to 2–93)
Otto Warburg was one of the pioneering scientists who elucidated the biochemical basis of glycolysis in the early 1900s. He also made a puzzling discovery about
cancer cells that remains interesting and relevant today. Normal diferentiated
cells, which do not proliferate, rely primarily upon aerobic respiration to generate
ATP, except when they experience decreased levels of oxygen, in which case they
switch to glycolysis. In contrast, cancer cells rely almost entirely upon glycolysis
for producing ATP, regardless of the presence of oxygen. Cancer cells can increase
the rate of glycolysis up to 200-fold relative to normal diferentiated cells. his
phenomenon, known as the Warburg efect, is still poorly understood. It is also
a focus of renewed interest because it suggests a unique attribute of cancer cells
that could be exploited to kill them selectively.
2–90
Which of the following properties of cancer cells would have provided
evidence for the Warburg efect?
(A) CROSS-FEEDING RESULT
TrpE–
TrpD–
TrpB–
(B) SYNTHETIC PATHWAY
chorismate
anthranilate
indole
tryptophan
Figure 2–22 Deining the pathway
for tryptophan synthesis using crossfeeding experiments (Problem 2–89).
(A) Results of a cross-feeding experiment
among mutants defective for steps in
the tryptophan biosynthetic pathway.
Orange areas on the Petri dish show
regions of heavy cell growth. (B) The
tryptophan biosynthetic pathway. Several
steps precede chorismate in the pathway
and there are several steps between
anthranilate and indole.
MCAT STYLE
I. Increased oxidation of pyruvate
II. Increased release of CO2
III. Increased release of lactate
A. I
B. II
C. III
D. I and II
2–91
You want to develop a drug that speciically kills cancer cells. Knowing about the Warburg efect, you hypothesize that cancer cells may be
especially sensitive to inhibitors of glycolysis because they are uniquely
dependent upon a high rate of glycolysis. Which one of the following
would be a good target for drug development?
A. Acetyl CoA
B. Isocitrate dehydrogenase
C. Pyruvate kinase
D. he electron-transport chain
2–92
Cancer cells undergo rapid unrestrained proliferation. he increased reliance of cancer cells upon glycolysis therefore seems paradoxical, since
glycolysis generates only 2 ATP from each glucose molecule, whereas
complete oxidative respiration of glucose generates up to 36 ATPs per
molecule. Which of the following best explains why glycolysis may be
advantageous to rapidly proliferating cancer cells?
A. Glycolysis generates NADH that can be used as a source of reducing
power for macromolecular synthesis.
B. Glycolysis produces intermediates that can be used to generate macromolecules needed for cell growth.
C. Relative to aerobic respiration, glycolysis produces more NADH, which
can be used as an energy source.
D. Unlike aerobic respiration, glycolysis produces fatty acids, which can be
stored as a source of energy.
2–93
he Warburg efect is exploited in an imaging technique that is commonly used to detect tumors. An individual is dosed with a molecule
labeled with a radioactive isotope of luorine (18F). he labeled molecule
is preferentially taken up by cancer cells and is detected by positron
emission tomography (PET scanning). Which of the following molecules
would you label with 18F to detect tumors via the Warburg efect?
A. Acetyl CoA
B. Glucose
C. Lactate
D. Pyruvate
Passage 2 (Questions 2–94 to 2–95)
Extremophiles are microorganisms that can survive and proliferate in extreme
environments. One fascinating group of extremophiles, called lithotrophs, are
remarkable organisms that are found deep beneath the Earth’s surface, living on
rocks under anaerobic conditions and surviving on CO2 as their sole source of
carbon.
2–94
Imagine that you are studying a newly discovered lithotroph and are trying to determine what it uses as a source of electrons for reducing CO2
and for producing energy. Which one of the following conditions must
be met for a molecule to serve as a useful electron donor?
A. Oxidation of the molecule occurs with a decrease in free energy
B. Oxidation of the molecule occurs with an increase in free energy
C. Reduction of the molecule occurs with a decrease in free energy
D. Reduction of the molecule occurs with an increase in free energy
29
30
2–95
Chapter 2: Cell Chemistry and Bioenergetics
Which of the following is most likely to be used by the lithotrophs as a
source of electrons for reducing CO2 to form useful organic molecules
and as an energy source for generating ATP?
A. Oxidation of glucose
B. Oxidation of H2S
C. Reduction of H2
D. Reduction of NO2
Chapter 3
31
CHAPTER
3
Proteins
THE SHAPE AND STRUCTURE OF PROTEINS
TERMS TO LEARN
α helix
amyloid fibril
β sheet
binding site
coiled-coil
conformation
polypeptide backbone
primary structure
prion disease
protein
protein domain
protein subunit
quaternary structure
secondary structure
side chain
tertiary structure
DEFINITIONS
Match the deinition below with its term from the list above.
3–1
hree-dimensional relationship of the diferent polypeptide chains in a
multisubunit protein or protein complex.
3–2
Common folding pattern in proteins in which a linear sequence of amino
acids folds into a right-handed coil stabilized by internal hydrogenbonding between backbone atoms.
3–3
he amino acid sequence of a protein.
3–4
A region on the surface of a protein that can interact with another molecule through noncovalent bonding.
3–5
Self-propagating, stable aggregate made up of identical polypeptide
chains layered into a continuous stack of β sheets.
3–6
he chain of repeating carbon and nitrogen atoms, linked by peptide
bonds, in a protein.
3–7
Common structural motif in proteins in which diferent sections of the
polypeptide chain run alongside each other and are joined together by
hydrogen-bonding between atoms of the polypeptide backbone.
3–8
Portion of a protein that has a tertiary structure of its own.
3–9
Regular local folding patterns in a protein, including α helix and β sheet.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
3–10
Each strand in a β sheet is a helix with two amino acids per turn.
3–11
Intrinsically disordered regions of proteins can be identiied using bioinformatic methods to search genes for encoded amino acid sequences
that possess high hydrophobicity and low net charge.
3–12
Loops of polypeptide that protrude from the surface of a protein often
form the binding sites for other molecules.
IN THIS CHAPTER
THE SHAPE AND STRUCTURE
OF PROTEINS
PROTEIN FUNCTION
32
Chapter 3: Proteins
3–13
Prion diseases can spread from one organism to another, providing that
the second organism eats tissue that contains the gene encoding the protein involved in formation of amyloid ibrils.
3–14
Why do you suppose that only L-amino acids and not a random mixture
of L- and D-amino acids are used to make proteins?
3–15
When egg white is heated, it hardens. his cooking process cannot be
reversed, but hard-boiled egg white can be dissolved by heating it in a
solution containing a strong detergent (such as sodium dodecyl sulfate)
together with a reducing agent, like 2-mercaptoethanol. Neither reagent
alone has any efect.
A. Why does boiling an egg white cause it to harden?
B. Why does it require both a detergent and a reducing agent to dissolve the
hard-boiled egg white?
3–16
Although α helices are common components of polypeptide chains, they
need to be of a certain minimum length. To ind out how chain length
afects α-helix formation, you measure the circular dichroism (a measure
of helicity) for a series of peptides of increasing length (Figure 3–1). Why
is there essentially no helix formation until the chain is at least six amino
acids long?
3–17
he uniform arrangement of the backbone carbonyl oxygens and amide
nitrogens in an α helix gives the helix a net dipole, so that it carries a partial positive charge at the amino end and a partial negative charge at the
carboxyl end. Where would you expect the ends of α helices to be located
in a protein? Why?
3–18
α Helices are often embedded in a protein so that one side faces the sur-
helicity
THOUGHT PROBLEMS
0
2
4
6
8
number of residues
10
Figure 3–1 Helicity of various peptides of
increasing length (Problem 3–16).
face and the other side faces the interior. Such helices are often termed
amphiphilic because the surface side is hydrophilic and the interior side
is hydrophobic. A simple way to decide whether a sequence of amino
acids might form an amphiphilic helix is to arrange the amino acids
around what is known as a “helix-wheel projection” (Figure 3–2). If the
hydrophobic and hydrophilic amino acids are segregated on opposite
sides of the wheel, the helix is amphiphilic. Using the helix-wheel projection, decide which of the three peptides in Figure 3–2 might form an
amphiphilic helix. (he mnemonic “FAMILY VW” will help you recognize
hydrophobic amino acids.)
3–19
Examine the segment of β sheet shown in Figure 3–3. For each strand of
the sheet decide whether it is parallel or antiparallel to each of its neighbors.
3–20
Like α helices, β sheets often have one side facing the surface of the
protein and one side facing the interior, giving rise to an amphiphilic
(A) HELIX WHEEL
15
8
1
(B) PEPTIDE SEQUENCES
12
4
5
S L
I K S V
I
E M V D E W F
F L
I R V L
R K V F
R L
F R S
R T
F
L
16
11
18
9
7
2
14
13
3
10
17
6
R V L K
I
R V L
T
R
I
L
S
A V
R
F
L
L
I
I
Figure 3–2 Helix-wheel projection (Problem
3–18). (A) Helix wheel. The circle (wheel)
represents the helix as viewed from one
end. Numbers show the positions of the
amino acid side chains, as projected
on the wheel. The positions of the irst
18 amino acids are shown; amino acid
19 would occupy the same position as
amino acid 1. Amino acid 1 is closest to
the reader; amino acid 18 is farthest
away. (B) Peptide sequences. The
N-termini are shown at the left;
hydrophobic amino acids are highlighted
in yellow; hydrophilic amino acids are
unmarked. (See Table 8, page 966, for
one-letter amino acid code.)
THE SHAPE AND STRUCTURE OF PROTEINS
33
sheet with one hydrophobic surface and one hydrophilic surface. From
the sequences listed below, pick the one that could form a strand in an
amphiphilic β sheet. hink about the way side chains are arranged in a
strand of a β sheet. (See Table 8, page 966, for one-letter amino acid code;
the mnemonic in Problem 3–18 might also be helpful.)
A. A L S C D V E T Y W L I
B. D K L V T S I A R E F M
C. D S E T K N A V F L I L
D. T L N I S F Q M E L D V
E. V L E F M D I A S V L D
3–21
Several diferent protein folds are represented in schematic form in
Figure 3–4. hese diagrams preserve the topology of the protein and
allow one to decide, for example, whether a protein is folded in a new
way or is an example of a protein fold that is already known. hese diagrams also permit a ready demonstration of a fundamental principle of
protein folding. For each of these folds, imagine that you could grasp the
N- and C-termini and pull them apart. Would any of the illustrated folds
produce a knot when fully stretched out?
3–22
It is a common observation that antiparallel strands in a β sheet are connected by short loops, but that parallel strands are connected by α helices. Why do you think this is?
3–23
In 1968, Cyrus Levinthal pointed out a complication in protein folding
that is widely known as the Levinthal paradox. He argued that because
there are astronomical numbers of conformations open to a protein in
the denatured state, it would take a very long time for a protein to search
through all the possibilities to ind the correct one, even if it tested each
possible conformation exceedingly rapidly. Yet denatured proteins typically take less than a second to fold inside the cell or in the test tube. How
do you suppose that proteins manage to fold so quickly?
3–24
Comparison of a homeodomain protein from yeast and from Drosophila
shows that only 17 of 60 amino acids are identical. How is it possible for
a protein to change over 70% of its amino acids and still fold in the same
way?
3–25
Often, the hard part of protein structure determination by x-ray difraction is getting good crystals. In diicult cases, there are two common
approaches for obtaining crystals: (1) using fragments of the protein and
(2) trying homologous proteins from diferent species.
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
Figure 3–3 A segment of β sheet from
the interior of thioredoxin (Problem 3–19).
Amide nitrogens are indicated by circled Ns;
hydrogen bonds are shown as red lines.
Figure 3–4 Topological representations
of several protein folds (Problem 3–21).
Vertical red arrows represent strands in
β sheets; blue connectors may be loops
or helices. Thick blue diagonal lines are
above the plane of the page; thin blue
lines lie below the plane of the page.
34
Chapter 3: Proteins
Figure 3–5 Catabolite activator protein
from E. coli (Problem 3–25). Yellow
shading indicates its domain structure.
β7
C
β6
N
β1
β5
β2
β4
β3
Figure 3–6 The kelch repeat domain of
galactose oxidase from D. dendroides
(Problem 3–27). The seven individual
β propellers are color coded and labeled.
The N- and C-termini are indicated by
N and C.
A. Examine the protein in Figure 3–5. Where would you cleave this protein
to obtain fragments that might be expected to fold properly and perhaps
form crystals?
B. Why do you suppose that homologous proteins from diferent species
might difer in their ability to form crystals?
3–26
A common strategy for identifying distantly related homologous proteins
is to search the database using a short signature sequence indicative of
the particular protein function. Why is it better to search with a short
sequence than with a long sequence? Don’t you have more chances for a
“hit” in the database with a long sequence?
3–27
he so-called kelch motif consists of a four-stranded β sheet, which forms
what is known as a β propeller. It is usually found to be repeated four to
seven times, forming a kelch repeat domain in a multidomain protein.
One such kelch repeat domain is shown in Figure 3–6. Would you classify
this domain as an “in-line” or “plug-in” type domain?
3–28
Examine the three protein monomers in Figure 3–7. From the arrangement of complementary binding surfaces, which are indicated by similarly shaped protrusions and invaginations, decide which monomer
would assemble into a ring, which would assemble into a chain, and
which would assemble into a sheet.
3–29
Cro is a bacterial gene regulatory protein that binds to DNA to turn genes
of. It is a symmetrical “head-to-head” dimer. Each of the two subunits of
the dimer recognizes a particular short sequence of nucleotides in DNA.
If the sequence of nucleotides recognized by one subunit is represented
as an arrow (→), so that the “head” of the arrow corresponds to DNA recognized by the “head” of the subunit, which of the following sequences
in DNA represents the binding site for the Cro dimer?
A. →→
B. →←
C. ←←
D. ←→
E. Could be more than one of the above
3–30
Why is it that there are numerous examples of “head-to-head” and “tailto-tail” dimers, but few, if any, examples of “head-to-tail” dimers?
(A)
(B)
(C)
Figure 3–7 Three protein monomers
(Problem 3–28).
THE SHAPE AND STRUCTURE OF PROTEINS
35
+
–
+
d
+
a
+
–
+
d
+
a
d
a
a
–
–
d
a
d
–
a
d
–
1
2
3
4
5
Figure 3–8 Binding surfaces for six
different proteins (Problem 3–31). In each
case, the bulk of the protein is below the
plane of the page.
6
3–31
Proteins bind to one another via weak interactions across complementary surfaces. Oppositely charged amino acids are apposed, as are hydrogen-bond donors and acceptors, and protrusions match invaginations
so that van der Waals contacts can be maximized. When two copies of
a protein bind to form a “head-to-head” dimer, they use the same binding surface. Examine the binding surfaces of the six proteins shown in
Figure 3–8, where charged amino acids are indicated by + and –, and
hydrogen-bond donors and acceptors are indicated by d and a. (Protrusions and invaginations—three-dimensional shapes—are not represented in the binding surfaces in Figure 3–8 just because it is diicult to
do so, but their absence does not change the general principles derived
from this problem.) In which cases could two copies of one protein form
a “head-to-head” dimer in which the charges and hydrogen-bonding
groups are appropriately matched? Can you spot any common feature of
the surfaces that allows such dimers to form?
3–32
Nuclear lamin C is a member of the intermediate ilament family. hus,
it should show regions of the coiled-coil heptad repeat motif AbcDefg,
where A and D are hydrophobic amino acids and b, c, e, f, and g can be
almost any amino acid. he sequence of nuclear lamin C is shown in
Figure 3–9 with potential coiled-coil regions highlighted. Examine
the segment marked “coil 1A.” How well does it conform to the heptad
repeat? (Don’t forget the mnemonic FAMILY VW.)
coil 1A
METPSQRRATRSGAQASSTPLSPTRITRLQEKEDLQELNDRLAVYIDRVRSLETENA
coil 1B
GLRLRITESEEVVSREVSGIKAAYEAELGDARKTLDSVAKERARLQLELSKVREEFK
ELKARNTKKEGDLIAAQARLKDLEALLNSKEAALSTALSEKRTLEGELHDLRGQVAK
LEAALGEAKKQLQDEMLRRVDAENRLQTMKEELDFQKNIYSEELRETKRRHETRLVE
coil 2
IDNGKQREFESRLADALQQLRAQHEDQVEQYKKELEKTYSAKLDNARQSAERNSNLV
GAAHEELQQSRIRIDSLSAQLSQLQKQLAAKEAKLRDLEDSLARERDTSRRLLAEKE
REMAEMRARMQQQLDEYQQLLDIKLALDMQIHAYRKLLEGEEERLRLSPSPTSQRSR
GRASSHSSQTQGGGSVTKKRKLESTESRSSPSQHARTSGRVAVEEVDEEGKFVRLRN
KSNEDQSMGNWQIKRQNGDDPLLTYRFPPKFTLKAGQVVTIWAAGAGATHSPPTDLV
WKAQNTWGCGNSLRTALINSTGEEVAMRKLVRSVTVVEDDEDEDGDDLLHHHHVSGS
RR
CALCULATIONS
3–33
Typical proteins have a stability ranging from 30 to 60 kJ/mole at 37°C.
Stability is a measure of the equilibrium between the folded and unfolded
forms of the protein:
folded [F] ↔ unfolded [U], K = [U]/[F]
For a protein with a stability of 41.5 kJ/mole, calculate the fraction of
unfolded protein that would exist at equilibrium at 37°C. At equilibrium,
ΔG° = – RT lnK = – 2.3RT log K,
where R = 8.3 × 10–3 kJ/(mole K) and T is temperature in K (37°C = 310 K).
Figure 3–9 The amino acid sequence of
nuclear lamin C (Problem 3–32).
Chapter 3: Proteins
3–34
Consider the following statement. “To produce one molecule of each possible kind of polypeptide chain, 300 amino acids in length, would require
more atoms than exist in the universe.” Given the size of the universe,
do you suppose this statement could possibly be correct? Since counting
atoms is a tricky business, consider the problem from the standpoint of
mass. he mass of the observable universe is estimated to be about 1080
grams, give or take an order of magnitude or so. Assuming that the average mass of an amino acid is 110 daltons, what would be the mass of one
molecule of each possible kind of polypeptide chain 300 amino acids in
length? Is this greater than the mass of the universe?
DATA HANDLING
Most proteins denature at both high and low pH. At high pH, the ionization of internal tyrosines is thought to be the main destabilizing inluence, whereas at low pH, the protonation of buried histidines (Figure
3–10A) is the likely culprit. A titration curve for the unfolding of the
enzyme ribonuclease is shown in Figure 3–10B. Superimposed on it is
the expected titration curve for the ionization of a histidine side chain
with a pK of about 4, which is typical for a buried histidine (the pK for
the side chain of the free amino acid is 6). he titration curve for denaturation is clearly much steeper than that for the side chain. Given the
discrepancy between the titration curves for protein unfolding and histidine protonation, how can it be true that protonation of histidine causes
protein unfolding?
3–35
3–36
(A)
CH2
+ H
N
CH2
+
+HN
NH
NH
(B)
1
protein
unfolding
fraction unfolded
or protonated
36
0.5
histidine
protonation
0
2
3
4
5
pH
6
7
Figure 3–10 Denaturation of proteins
(Problem 3–35). (A) Histidine protonation.
(B) Titration curves for protein unfolding
and histone protonation.
Titin, which has a molecular weight of about 3 × 106, is the largest polypeptide yet described. Titin molecules extend from muscle thick ilaments to the Z disc; they are thought to act as springs to keep the thick
ilaments centered in the sarcomere. Titin is composed of a large number
of repeated immunoglobulin (Ig) sequences of 89 amino acids, each of
which is folded into a domain about 4 nm in length (Figure 3–11A).
You suspect that the springlike behavior of titin is caused by the
sequential unfolding (and refolding) of individual Ig domains. You test
this hypothesis using the atomic force microscope, which allows you
to pick up one end of a protein molecule and pull with an accurately
measured force. For a fragment of titin containing seven repeats of the
Ig domain, this experiment gives the sawtooth force-versus-extension
curve shown in Figure 3–11B. If the experiment is repeated in a solution
of 8 M urea (a protein denaturant), the peaks disappear and the measured extension becomes much longer for a given force. If the experiment
is repeated after the protein has been cross-linked by treatment with glutaraldehyde, once again the peaks disappear but the extension becomes
much smaller for a given force.
A. Are the data consistent with your hypothesis that titin’s springlike behavior is due to the sequential unfolding of individual Ig domains? Explain
your reasoning.
(A)
(B)
C
force (pN)
400
N
300
200
100
0
0
50
100
extension (nm)
150
200
8
Figure 3–11 Springlike behavior of
titin (Problem 3–36). (A) The structure
of an individual Ig domain. (B) Force
in piconewtons versus extension in
nanometers obtained by atomic force
microscopy.
PROTEIN FUNCTION
37
B. Is the extension for each putative domain-unfolding event the magnitude you would expect? (In an extended polypeptide chain, amino acids
are spaced at intervals of 0.34 nm.)
C. Why is each successive peak in Figure 3–11B a little higher than the one
before?
D. Why does the force collapse so abruptly after each peak?
3–37
You are skeptical of the blanket statement that cysteines in intracellular
proteins are not involved in disulide bonds, while in extracellular proteins they are. To test this statement you carry out the following experiment. As a source of intracellular protein you use reticulocytes, which
have no internal membranes and, thus, no proteins from the endoplasmic reticulum (ER) or other membrane-enclosed compartments. As
examples of extracellular proteins, you use bovine serum albumin (BSA),
which has 37 cysteines, and insulin, which has 6. You denature the soluble proteins from a reticulocyte lysate and the two extracellular proteins
so that all cysteines are exposed. To probe the status of cysteines, you
treat the proteins with N-ethylmaleimide (NEM), which reacts covalently with the –SH groups of free cysteines, but not with sulfur atoms in
disulide bonds. In the irst experiment, you treat the denatured proteins
with radiolabeled NEM, then break any disulide bonds with dithiothreitol (DTT) and react a second time with unlabeled NEM. In the second
experiment, you do the reverse: you irst treat the denatured proteins
with unlabeled NEM, then break disulide bonds with DTT and treat with
radiolabeled NEM. he proteins are separated according to size by electrophoresis on a polyacrylamide gel (Figure 3–12).
A. Do any cytosolic proteins have disulide bonds?
B. Do the extracellular proteins have any free cysteine –SH groups?
C. How do you suppose the results might difer if you used lysates of cells
that have internal membrane-enclosed compartments?
PROTEIN FUNCTION
TERMS TO LEARN
active site
allosteric protein
antibody
antigen
catalyst
coenzyme
equilibrium constant (K)
enzyme
feedback inhibition
GTP-binding protein
(GTPase)
ligand
linkage
lysozyme
motor protein
protein kinase
protein phosphatase
proteomics
regulatory site
scaffold protein
substrate
transition state
ubiquitin
ubiquitin ligase
DEFINITIONS
Match the deinition below with its term from the list above.
3–38
A protein that serves both to link together a set of interacting proteins
and to position them at a speciic location in a cell.
3–39
Type of metabolic regulation in which the activity of an enzyme acting
near the beginning of a reaction pathway is reduced by a product of the
pathway.
3–40
Protein produced by the immune system in response to a foreign molecule or invading microorganism.
3–41
Region of an enzyme surface to which a substrate molecule binds in
order to undergo a catalyzed reaction.
3–42
A protein catalyst that speeds up a reaction, often by a factor of a million
or more, without itself being changed.
lysate
insulin + BSA
*NEM
NEM
*NEM
DTT
DTT
DTT
NEM
DTT
NEM
*NEM
NEM
*NEM
albumin
insulin
chains
Figure 3–12 Test for disulide bonds
in cytosolic and extracellular proteins
(Problem 3–37). The order of treatment
with NEM and DTT is indicated at the top
of each lane; *NEM indicates radiolabeled
NEM.
38
Chapter 3: Proteins
3–43
he irst example of a special family of small proteins whose members
are covalently attached to other proteins to inluence their activity or fate.
3–44
Mutual efect of the binding of one ligand on the binding of another that
is a central feature of the behavior of all allosteric proteins.
3–45
Enzyme that transfers the terminal phosphate group of ATP to a speciic
amino acid in a target protein.
3–46
Rate-limiting structure that forms transiently in the course of a chemical
reaction and has the highest free energy of any reaction intermediate.
3–47
Protein that changes its conformation (and often its activity) when it
binds a regulatory molecule or when it is covalently modiied.
3–48
A term often used to describe research focused on the simultaneous
analysis of large numbers of proteins.
3–49
Small molecule that is tightly associated with a protein catalyst and participates in the chemical reaction, often by forming a covalent bond to
the substrate.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
3–50
he tendency for an amino acid side-chain group such as –COOH to
release a proton, its pK, is the same for the amino acid in solution and for
the amino acid in a protein.
3–51
For a family of related genes that do not match genes of known function
in the sequence database, it should be possible to deduce their function
by using “evolutionary tracing” to see where conserved amino acids cluster on their surfaces.
3–52
Higher concentrations of enzyme give rise to a higher turnover number.
3–53
Enzymes that undergo cooperative allosteric transitions invariably consist of symmetric assemblies of multiple subunits.
3–54
Continual addition and removal of phosphates by protein kinases and
protein phosphatases is wasteful of energy—since their combined action
consumes ATP—but it is a necessary consequence of efective regulation
by phosphorylation.
3–55
Conformational changes in proteins never exceed a few tenths of a
nanometer.
THOUGHT PROBLEMS
3–56
3–57
Antarctic notothenioid ish (Figure 3–13) avoid freezing in their perpetually icy environment because of an antifreeze protein that circulates in
their blood. his evolutionary adaptation has allowed the Notothenioidei
suborder to rise to dominance in the freezing Southern Ocean. It is said
that all proteins function by binding to other molecules. To what ligand
do you suppose antifreeze proteins bind to keep the ish from freezing?
Or do you think this might be an example of a protein that functions in
the absence of any molecular interaction?
How does the protein environment surrounding an amino acid side
chain afect its chemical properties? Consider the carboxyl group on an
aspartate side chain in the following environments in a protein. Rank
order these environments from the highest to the lowest proportion of
Dissostichus eleginoides,
the Chilean sea bass
Pagothenia borchgrevinki
Figure 3–13 Two notothenioid ish
(Problem 3–56). The notothenioid family
now dominates Antarctica’s continental
shelf, accounting for 50% of the species
and 95% of the biomass of ish. The
Chilean sea bass is commonly served in
restaurants.
PROTEIN FUNCTION
Aminoacyl-tRNA synthetases attach speciic amino acids to their appropriate tRNAs in preparation for protein synthesis. he synthetase that
attaches valine to tRNAVal must be able to discriminate valine from threonine, which difer only slightly in structure: valine has a methyl group
where threonine has a hydroxyl group (Figure 3–14). Valyl-tRNA synthetase achieves this discrimination in two steps. In the irst, it uses a
binding pocket whose contours allow valine or threonine (but not other
amino acids) to bind, but the binding of valine is preferred. his site is
responsible for coupling the amino acid to the tRNA. In the second step,
the enzyme checks the newly made aminoacyl-tRNA using a second
binding site that is very speciic for threonine and hydrolyzes it from the
tRNA. How do you suppose it is that the second binding site can be very
speciic for threonine, whereas the irst binding site has only a moderate
speciicity for valine?
3–59
You have raised a speciic, high-ainity monoclonal antibody against the
enzyme you are working on, and have identiied its interaction site as a
stretch of six amino acids in the enzyme. Your advisor suggests that you
could use the antibody to purify the enzyme by ainity chromatography.
his technique would involve attaching the antibody to the inert matrix
of a column, passing a crude cell lysate over the column, allowing the
antibody to bind your enzyme but not other proteins, and inally eluting your enzyme by washing the column with a solution containing the
six-amino-acid peptide corresponding to the binding site. he principal
advantage of ainity chromatography is that it allows a rapid, one-step
puriication under mild conditions that retain enzyme activity.
In a preliminary experiment, you show that if you incubate the antibody with the peptide corresponding to the binding site it will no longer
bind to your enzyme, demonstrating that the antibody binds the peptide.
Encouraged, you bind the antibody to the column and show that it completely removes your enzyme from the crude cell lysate. When you try
to elute your enzyme with a solution containing a high concentration of
the peptide, however, you ind that none of your enzyme comes of the
column. What could have gone wrong? (hink about what must happen
for the enzyme to come of the column.)
3–60
Examine Figure 3–15, which compares the energetics of a catalyzed and
uncatalyzed reaction during the progress of the reaction from substrate
(S) to product (P). he highest peak in such a diagram corresponds to the
transition state, which is an unstable, high-energy arrangement of substrate atoms that is intermediate between substrate and product. he free
energy required to surmount this barrier to the reaction is termed the
activation energy. Enzymes function by lowering the activation energy,
thereby allowing a more rapid approach to equilibrium.
With this diagram in mind, consider the following question. Suppose
the enzyme in the diagram were mutated in such a way that its ainity
for the substrate was increased by a factor of 100. Assume that there was
no other efect beyond increasing the depth of the trough labeled ES
CH3
H3C
OH
CH
CH
CH
CH
COO–
+H N
3
+H N
3
valine
COO–
threonine
Figure 3–14 Structures of valine and
threonine (Problem 3–58).
UNCATALYZED
free energy
3–58
H3C
activation
energy
S
P
progress of reaction
CATALYZED
free energy
carboxyl groups in the –COO– form; that is, in terms of their pKas. Explain
your ranking.
1. An aspartate side chain on the surface of a protein with no other ionizable groups nearby.
2. An aspartate side chain buried in a hydrophobic pocket on the surface of
a protein.
3. An aspartate side chain in a hydrophobic pocket adjacent to a glutamate
side chain.
4. An aspartate side chain in a hydrophobic pocket adjacent to a lysine side
chain.
39
activation
energy
S
ES
P
progress of reaction
Figure 3–15 Catalyzed and uncatalyzed
reactions showing the free energy at
various stages in the progress of the
reaction (Problem 3–60).
40
Chapter 3: Proteins
(enzyme–substrate complex) in Figure 3–15. Would you expect the rate
of the reaction catalyzed by the altered enzyme to be faster, slower, or
equal to the reaction rate catalyzed by the normal enzyme?
3–61
Which one of the following properties of an enzyme is responsible for
its saturation behavior; that is, a maximum rate insensitive to further
increases in substrate concentration?
A. he enzyme does not change the overall equilibrium constant for a reaction.
B. he enzyme lowers the activation energy of a chemical reaction.
C. he enzyme is a catalyst that is not consumed by the reaction.
D. he enzyme has a ixed number of active sites where substrate binds.
E. he product of the enzyme reaction usually inhibits the enzyme.
3–62
he Michaelis constant, Km, is often spoken of as if it were a measure of
the ainity of the enzyme for the substrate: the lower the Km, the higher
the binding ainity. his would be true if Km were the same as Kd (the
equilibrium constant for the dissociation reaction), but it is not. For an
enzyme-catalyzed reaction
E+S
Km =
k1
kcat
ES → E + P
k–1
(k–1 + kcat)
k1
A. In terms of these rate constants, what is Kd for dissociation of the ES complex to E + S?
B. Under what conditions is Km approximately equal to Kd?
C. Does Km consistently overestimate or underestimate the binding ainity? Or does it sometimes overestimate and sometimes underestimate
the binding ainity?
3–63
You are trying to determine whether it is better to purify an enzyme from
its natural source or to express the gene in bacteria and then purify it. You
purify the enzyme in the same way from both sources and show that each
preparation gives a single band by denaturing gel electrophoresis, a common measure of purity. When you compare the kinetic parameters, you
ind that both enzymes have the same Km but the enzyme from bacteria
has a 10-fold lower Vmax. Propose possible explanations for this result.
3–64
he enzyme hexokinase adds a phosphate to D-glucose but ignores its
mirror image, L-glucose. Suppose that you were able to synthesize hexokinase entirely from D-amino acids, which are the mirror image of the
normal L-amino acids.
A. Assuming that the “D” enzyme would fold to a stable conformation, what
relationship would you expect it to bear to the normal “L” enzyme?
B. Do you suppose the “D” enzyme would add a phosphate to L-glucose,
and ignore D-glucose?
3–65
In 1948, Linus Pauling proposed what is now considered to be a key
aspect of enzyme function.
“I believe that an enzyme has a structure closely similar to that found
for antibodies, but with one important diference, namely, that the surface coniguration of the enzyme is not so closely complementary to its
speciic substrate as is that on an antibody, but is instead complementary to an unstable molecule with only transient existence—namely, the
‘activated complex’ [transition state, in modern parlance] for the reaction that is catalyzed by the enzyme. he mode of action of an enzyme
would then be the following: the enzyme would show a small power of
PROTEIN FUNCTION
41
H
H
H
C
OH
C
OH
C
O
C
O–
H2 C
OPO32–
H2 C
dihydroxyacetone
phosphate
Figure 3–16 The reaction catalyzed by
triosephosphate isomerase, and the
enzyme inhibitor, phosphoglycolate
(Problem 3–65).
H
H
OPO32–
C
O
C
OH
H2 C
cis-enediolate
intermediate
OPO32–
glyceraldehyde
3-phosphate
O
C
H2 C
O–
OPO32–
phosphoglycolate
inhibitor
attraction for the substrate molecule or molecules, which would become
attached to it in its active surface region. his substrate molecule, or
these molecules, would then be strained by the forces of attraction to
the enzyme, which would tend to deform it into the coniguration for the
activated complex, for which the power of attraction by the enzyme is the
greatest…. he assumption made above that the enzyme has a coniguration complementary to the activated complex, and accordingly has the
strongest power of attraction for the activated complex, means that the
activation energy for the reaction is less in the presence of the enzyme
than in its absence, and accordingly that the reaction would be speeded
up by the enzyme.”
he enzyme triosephosphate isomerase catalyzes the interconversion of glyceraldehyde 3-phosphate and dihydroxyacetone phosphate
through a cis-enediolate intermediate (Figure 3–16). Phosphoglycolate
(Figure 3–16) is a competitive inhibitor of triosephosphate isomerase
with a Kd of 7 μM. he normal substrates for the enzyme have a Kd of
about 100 μM. Do you think that phosphoglycolate is a transition-state
analog? Why or why not?
3–66
he mechanism for lysozyme cleavage of its polysaccharide substrate
requires Glu35 in its nonionized form, whereas the nearby Asp52 must be
ionized (Figure 3–17). he pK values for the side-chain carboxyl groups
on the two amino acids in solution are virtually identical.
A. How can one carboxyl group be charged and the other uncharged in the
active site of lysozyme?
B. he pH optimum for lysozyme is about 5. Why do you suppose that the
activity decreases above and below this optimum?
Glu35
NAG
O
C
Asp52
O
Glu35
H
O
C
C
O
–O
O
C
O
tri-NAG
NAG
Asp52
Figure 3–17 Forms of Glu35 and Asp52
required for polysaccharide cleavage by
lysozyme (Problem 3–66). In the spaceilling model, the positions of Glu35
and Asp52 are shown relative to the
trisaccharide of N-acetylglucosamine
(NAG) units, tri-NAG, which is not
quite long enough to be cleaved. In
the schematic diagram, the positions
of Glu35 and Asp52 are shown relative
to the glycosidic bond to be cleaved
in a polysaccharide composed of NAG
residues.
42
Chapter 3: Proteins
R5P
A
B
C
D
F
G
AMP
H
I
GMP
E
3–67
How do you suppose that a molecule of hemoglobin is able to bind oxygen eiciently in the lungs, and yet release it eiciently in the tissues?
3–68
If you were in charge of enzyme design for a cell, for what circumstances
might you design an enzyme that had a Km much, much lower than the
prevailing substrate concentration ([S] >> Km)? A Km around the prevailing substrate concentration ([S] ≈ Km)? A Km much, much higher than
the prevailing substrate concentration ([S] << Km)?
3–69
Which of the following does NOT describe a mechanism that cells use to
regulate enzyme activities?
A. Cells control enzyme activity by phosphorylation and dephosphorylation.
B. Cells control enzyme activity by the binding of small molecules.
C. Cells control the rates of difusion of substrates to enzymes.
D. Cells control the rates of enzyme degradation.
E. Cells control the rates of enzyme synthesis.
F. Cells control the targeting of enzymes to speciic organelles.
3–70
Synthesis of the purine nucleotides AMP and GMP proceeds by a
branched pathway starting with ribose 5-phosphate (R5P), as shown
schematically in Figure 3–18. Using the principles of feedback inhibition,
propose a regulatory strategy for this pathway that ensures an adequate
supply of both AMP and GMP and minimizes the buildup of the intermediates (A–I) when supplies of AMP and GMP are adequate.
3–71
Pathways devoted to the synthesis of speciic bioproducts such as purines
are commonly regulated via feedback inhibition by the inal product. By
contrast, the low of metabolites through the web of pathways devoted
to overall energy metabolism—production and utilization of ATP, as well
as the buildup and breakdown of internal fuel reserves—is regulated by
metabolites whose concentrations relect the energy status of the cell.
ATP-like signal metabolites (such as ATP and NADH) tend to accumulate
when the cell is slowly consuming ATP to meet its energy needs; AMPlike signal metabolites (such as AMP, ADP, inorganic phosphate, and
NAD+) tend to accumulate when the cell is rapidly using ATP.
Consider the pathways for the synthesis and breakdown of glycogen,
the main fuel reserve in muscle cells (Figure 3–19). he synthetic pathway is controlled by glycogen synthase, whereas the breakdown pathway
is controlled by glycogen phosphorylase. In resting muscle, which type of
signal metabolite would be expected to accumulate? How would those
signal metabolites be expected to afect the activity of the two regulated
enzymes of glycogen metabolism? What about in exercising muscle?
Figure 3–18 Schematic diagram of the
metabolic pathway for synthesis of AMP
and GMP from R5P (Problem 3–70).
glycogen
phosphorylase
ATP
GLYCOGEN
glycogen
synthase
G1P
G6P
UDPG
GLUCOSE
ATP
CO2
Figure 3–19 Pathways for glycogen
synthesis and its breakdown to glucose
6-phosphate, which is an intermediate
along the pathway for glucose
metabolism to Co2 (Problem 3–71). G6P
stands for glucose 6-phosphate, G1P
for glucose 1-phosphate, and UDPG for
uridine diphosphoglucose.
PROTEIN FUNCTION
3–72
he enzyme glycogen phosphorylase uses phosphate as a substrate to
split of glucose 1-phosphate from glycogen, which is a polymer of glucose. Glycogen phosphorylase in the absence of any ligands is a dimer
that exists in two conformations: a predominant one with low enzymatic
activity and a rarer one with high activity. Both phosphate, a substrate
that binds to the active site, and AMP, an activator that binds to an allosteric site, alter the conformational equilibrium by binding preferentially
to one conformation. To which conformation of the enzyme would you
expect phosphate to bind, and why? To which conformation would you
expect AMP to bind, and why? How does the binding of either molecule
alter the activity of the enzyme?
3–73
Monod, Wyman, and Changeux (MWC) originally explained the kinetic
behavior of allosteric enzymes using four postulates, as summarized
below.
1. All subunits in the allosteric enzyme have identical conformations and
are arranged symmetrically.
2. Each subunit carries a binding site for each ligand.
3. he allosteric enzyme can exist in at least two conformations that conserve its overall symmetry. he diferent conformations may have very
diferent ainities for the ligands, which may be bound to any combination of subunits.
4. he binding ainity of a ligand depends only on the conformational state
of the allosteric enzyme and not on the occupancy of neighboring sites.
A subset of all the arrangements of subunits in an allosteric enzyme
composed of four identical subunits is shown in Figure 3–20. Each subunit has two possible conformations and dark blue subunits have a bound
ligand. Assuming that the ligand binds much more tightly to one conformation of subunit (circle), decide which of the tetrameric species are
consistent with the MWC postulates. What would your answer be if the
ligand bound equally well to each of the two conformations of subunit?
3–74
Many proteins inside cells are regulated by phosphorylation and dephosphorylation. Consider a metabolic reaction catalyzed by an enzyme that
is fully active when not phosphorylated and completely inactive when it
is phosphorylated (Figure 3–21). Inside the cell, the rate of this metabolic
reaction can vary continuously from very fast to very slow (or zero) at any
given substrate concentration. How is it that the reaction in the cell can
proceed at any rate between very fast and zero, even though individual
enzyme molecules are either fully on or completely turned of?
3–75
Motor proteins generally require ATP (or GTP) hydrolysis to ensure unidirectional movement.
A. In the absence of ATP, would you expect a motor protein to stop moving,
to wander back and forth, to move in reverse, or to continue moving forward but more slowly?
B. Assume that the concentrations of ATP, ADP, and phosphate were
adjusted so that the free-energy change for ATP hydrolysis by the motor
protein was equal to zero (instead of very negative, as it is normally).
Under these conditions, would you expect a motor protein to stop moving, to wander back and forth, to move in reverse, or to continue moving
forward but more slowly?
43
Figure 3–20 A subset of all the possible
arrangements of a tetramer, composed of
subunits with either of two conformations
(Problem 3–73). Circles and squares
represent the two conformations of the
subunits; dark blue indicates a subunit
with a bound ligand.
PO42–
CALCULATIONS
ON
3–76
An antibody binds to another protein with an equilibrium constant, K,
of 5 × 109 M–1. When it binds to a second, related protein, it forms three
fewer hydrogen bonds, reducing its binding ainity by 11.9 kJ/mole.
OFF
Figure 3–21 Phosphorylated and
nonphosphorylated states of a metabolic
enzyme (Problem 3–74).
44
Chapter 3: Proteins
What is the K for its binding to the second protein? (Free-energy change
is related to the equilibrium constant by the equation ΔG° = –2.3 RT log
K, where R is 8.3 × 10–3 kJ/(mole K) and T is 310 K.)
3–77
he equilibrium constant for a reaction like that of antibody (Ab) binding
to a protein (Pr) to form an antibody–protein complex (Ab–Pr) is equal
to the ratio of the association rate constant, kon, to the dissociation rate
constant, kof (K = [Ab–Pr]/([Ab][Pr]) = kon/kof). Recall that the association rate (kon[Ab][Pr]) equals the dissociation rate (kof[Ab–Pr]) at equilibrium. Consider two such reactions. he irst has an on rate constant of
105 M–1 sec–1 and an of rate constant of 10–3 sec–1 at 37°C. he second has
an on rate constant of 103 M–1 sec–1 and an of rate of 10–5 sec–1 at 37°C.
A. What are the equilibrium constants for these two reactions?
B. At equal concentrations of antibody and protein, which of these reactions will reach its equilibrium point more quickly?
C. You wish to use an antibody to purify the protein you are studying. You
are concerned that the complex may fall apart in the time it takes you to
isolate it, and you are unsure how the of rates relate to the half-time for
dissociation; that is, the time at which half the complex will have dissociated. It can be shown that the fraction of complex remaining at time t
([Ab–Pr]t) relative to that present initially ([Ab–Pr]0) is
[Ab–Pr]t
= e–kofft
[Ab–Pr]0
an equation more easily dealt with in its logarithmic form,
2.3 log
[Ab–Pr]t
= –kofft
[Ab–Pr]0
Using this relationship, decide how long it will take for half of the Ab–Pr
complexes in each of the above two reactions to dissociate. Neglect any
contribution from new complex being formed in the association reaction.
3–78
Consider an uncatalyzed reaction, A
B. he rate constants for the forward and reverse reactions are kf = 10–4 sec–1 and kr = 10–7 sec–1. hus, the
rates or velocities (v) of the forward and reverse reactions are
vf = kf [A]
and
vr = kr [B]
he overall reaction rate is
v = vf – vr = kf [A] – kr [B]
A. What is the overall reaction rate at equilibrium?
B. What is the value of the equilibrium constant, K?
C. You now add an enzyme that increases kf by a factor of 109. What will the
value of the equilibrium constant be with the enzyme present? What will
the value of kr be?
3–79
Many enzymes obey simple Michaelis–Menten kinetics, which are summarized by the equation
rate =
Vmax [S]
[S] + Km
where Vmax = maximum velocity, [S] = concentration of substrate, and
Km = the Michaelis constant.
It is instructive to plug a few values of [S] into the equation to see how
rate is afected. What are the rates for [S] equal to zero, equal to Km, and
equal to ininite concentration?
3–80
Suppose that the enzyme you are studying is regulated by phosphorylation. When the enzyme is phosphorylated, the Km for its substrate
PROTEIN FUNCTION
45
increases by a factor of 3, but Vmax is unaltered. At a concentration of substrate equal to the Km for the unphosphorylated enzyme, decide whether
phosphorylation activates the enzyme or inhibits it? Explain your reasoning.
3–81
For an enzyme that follows Michaelis–Menten kinetics, by what factor
does the substrate concentration have to increase to change the rate of
the reaction from 20% to 80% Vmax?
A. A factor of 2
B. A factor of 4
C. A factor of 8
D. A factor of 16
E. he factor required cannot be calculated without knowing Km
3–82
he “turnover number,” or kcat, for an enzyme is the number of substrate
molecules converted into product by an enzyme molecule per unit time
when the enzyme is fully saturated with substrate. he maximum rate of
a reaction, Vmax, equals kcat times the concentration of enzyme. (Remember that the maximum rate occurs when all of the enzyme is present as
the ES complex.) Carbonic anhydrase catalyzes the hydration of CO2
to form H2CO3. Operating at its maximum rate, 10 μg of pure carbonic
anhydrase (Mr 30,000) in 1 mL hydrates 0.90 g of CO2 in 1 minute. What is
the turnover number for carbonic anhydrase?
3–83
Rous sarcoma virus (RSV) carries an oncogene called Src, which encodes
a continuously active protein tyrosine kinase that leads to unchecked
cell proliferation. Normally, Src carries an attached fatty acid (myristoylate) group that allows it to bind to the cytoplasmic side of the plasma
membrane. A mutant version of Src that does not allow attachment of
myristoylate does not bind to the membrane. Infection of cells with RSV
encoding either the normal or the mutant form of Src leads to the same
high level of protein tyrosine kinase activity, but the mutant Src does not
cause cell proliferation.
A. Assuming that the normal Src is all bound to the plasma membrane
and that the mutant Src is distributed throughout the cytoplasm, calculate their relative concentrations in the neighborhood of the plasma
membrane. For the purposes of this calculation, assume that the cell is
a sphere with a radius (r) of 10 μm and that the mutant Src is distributed
throughout the cell, whereas the normal Src is conined to a 4-nm-thick
layer immediately beneath the membrane. [For this problem, assume
that the membrane has no thickness. he volume of a sphere is (4/3)πr3.]
B. he target (X) for phosphorylation by Src resides in the membrane.
Explain why the mutant Src does not cause cell proliferation.
DATA HANDLING
3–84
he binding of platelet-derived growth factor (PDGF) to the PDGF
receptor stimulates phosphorylation of eight tyrosines in the receptor’s
cytoplasmic domain. he enzyme phosphatidylinositol 3ʹ-kinase (PI
3-kinase) binds to one or more of the phosphotyrosines through its SH2
domains and is thereby activated. To identify the activating phosphotyrosines, you synthesize eight pentapeptides that contain the critical
tyrosines (at the N-terminus) in their phosphorylated or unphosphorylated forms. You then mix an excess of each of the various pentapeptides with phosphorylated PDGF receptor and PI 3-kinase. Immunoprecipitation of the PDGF receptor will bring down any bound
PI 3-kinase, which can be assayed by its ability to add 32P-phosphate to
its substrate (Figure 3–22).
A. Do your results support the notion that PI 3-kinase binds to phosphotyrosines in the activated PDGF receptor? Why or why not?
pentapeptides with tyrosine
684 708 719 731 739 743 746 755
pentapeptides with phosphotyrosine
684 708 719 731 739 743 746 755
Figure 3–22 Assay for PI 3-kinase in
immunoprecipitates of the PDGF receptor
(Problem 3–84). Pentapeptides are
indicated by numbers that refer to the
position of tyrosine in the PDGF receptor.
Red circles indicate incorporation of
32P-phosphate into the substrate for
PI 3-kinase.
46
Chapter 3: Proteins
B. he amino acid sequences of the PDGF receptor pentapeptides tested
above (numbered according to the position of tyrosine in the PDGF
receptor) and of peptide segments that are known to bind to PI 3-kinase
in other activated receptors are shown below.
684 YSNAL
YMMMR (FGF receptor)
708 YMDMS
YTHMN (insulin receptor)
719 YVPML
YEVML (hepatocyte growth factor receptor)
731 YADIE
YMDMK (steel factor receptor)
739 YMAPY
YVEMR (CSF-1 receptor)
743 YDNYE
746 YEPSA
755 YRATL
What are the common features of peptide segments that form binding
sites for PI 3-kinase?
C. Which of the three common types of protein–protein interaction—
surface–string, helix–helix, or surface–surface—does the binding of
PI 3-kinase with the PDGF receptor most likely illustrate?
3–85
A common method for determining the equilibrium for binding of a
ligand (L) to a protein (Pr) is to use gel electrophoresis to separate the
bound (Pr–L) and free forms of the ligand. By convention, the equilibrium is usually considered for the dissociation reaction (Pr–L → Pr + L),
rather than the association reaction (Pr + L → Pr–L), and thus the equilibrium constant is referred to as the dissociation constant, Kd. his method
was used to show that the protein, SmpB, binds speciically to a special
species of tRNA (tmRNA) that is used in bacteria to eliminate the incomplete proteins made from truncated mRNAs. In this experiment, tmRNA
was labeled and included in the binding reactions at a concentration of
0.1 nM (10–10 M). Puriied SmpB protein was included at a range of concentrations and the mixture was incubated until the binding reaction was
at equilibrium. Free and bound tmRNA were then separated by electrophoresis and made visible by autoradiography (Figure 3–23). (Note that
this method requires that the of rate be slow enough that the complex
will not dissociate during the time it takes to carry out the electrophoresis, usually a few hours.)
A. Consider the equation Kd = [Pr][L]/[Pr–L]. When the concentrations of
bound and free ligand are equal, what is the relationship between the
concentration of free protein and the Kd?
B. By visual inspection of Figure 3–23, estimate the Kd. Do you have to worry
about the concentration of bound protein in this experiment? Why or
why not?
C. he concentration of labeled tmRNA in these experiments was 100 pM.
Would the results have been the same if tmRNA had been used at
100 nM? At 100 μM?
3–86
If the data in Figure 3–23 are plotted as fraction tmRNA bound versus
SmpB concentration, one obtains a symmetrical S-shaped curve as
shown in Figure 3–24. his curve is a visual display of a very useful relationship between Kd and concentration, which has broad applicability.
he expression for fraction of ligand bound is derived from the equation
bound
free
0.08
0.3
1.2
4.7
18.8
75
300 1200
concentration of SmpB (nM)
Figure 3–23 Assay of the binding of
puriied SmpB protein to 32P-labeled
tmRNA (Problem 3–85). From left to right
across the gel, the experiment in each
successive lane used a 2-fold increase
in concentration of SmpB protein;
concentrations in every other lane are
indicated.
PROTEIN FUNCTION
47
Figure 3–24 Fraction of tmRNA bound
versus SmpB concentration (Problem 3–86).
fraction bound
1.0
0.75
TABLE 3–1 Fraction of
ligand bound versus protein
concentration (Problem 3–86).
Protein
concentration
0.5
Fraction bound
(%)
104 Kd
0.25
103 Kd
0
10–11
10–9
10–7
10–5
concentration of SmpB (M)
for Kd by substituting ([L]TOT – [L]) for [Pr–L] and rearranging. Because
the total concentration of ligand ([L]TOT) is equal to the free ligand ([L])
plus bound ligand ([Pr–L]),
fraction bound = [Pr–L] = [Pr]
[L]TOT [Pr] + Kd
(An equivalent relationship in terms of the fraction of protein bound can
be derived in an analogous way; it is [Pr–L]/[Pr]TOT = [L]/([L] + Kd).)
Using this relationship, calculate the fraction of ligand bound for protein concentrations expressed in terms of Kd, using Table 3–1.
3–87
he rates of production of product, P, from substrate, S, catalyzed by
enzyme, E, were measured under conditions in which very little product
was formed. he results are summarized in Table 3–2.
A. Why is it important to measure rates of product formation under conditions in which very little product is formed?
B. Plot these data as rate versus substrate concentration. Is this plot a rectangular hyperbola as expected for an enzyme that obeys Michaelis–
Menten kinetics? What would you estimate as the Km and Vmax values for
this enzyme?
C. To obtain more accurate values for the kinetic constants, the Lineweaver–
Burk transformation of the Michaelis–Menten equation is often used so
that the data can be plotted as a straight line.
Michaelis–Menten equation:
rate =
101 Kd
Kd
10–1 Kd
10–2 Kd
10–3 Kd
10–4 Kd
TABLE 3–2 Initial rates of product
formation at various substrate
concentrations (Problem 3–87).
Rate
(μmol/min)
[S]
(μM)
0.15
0.08
0.21
0.12
0.7
0.54
1.1
1.23
Vmax [S]
1.3
1.82
[S] + Km
1.5
2.72
1.7
4.94
1.8
10.00
1/Rate
(min/μmol)
1/[S]
(1/μM)
6.7
12.5
4.8
8.3
1.4
1.9
0.91
0.81
0.77
0.55
0.67
0.37
0.59
0.20
0.56
0.10
Lineweaver–Burk equation:
1
Km 1
1
=
+
rate Vmax [S] Vmax
his equation has the form of a straight line, y = ax + b. hus, when
1/rate (y) is plotted versus 1/[S] (x), the slope of the line equals Km/Vmax
(a) and the y intercept is 1/Vmax (b). Furthermore, it can be shown that
the x intercept is equal to –1/Km.
Plot 1/rate versus 1/[S] and determine the kinetic parameters Km and
Vmax. (he values for 1/rate and 1/[S] are shown in Table 3–2.)
3–88
102 Kd
Lysozyme achieves its antibacterial efect by cleaving the polysaccharide chains that form the bacterial cell wall. In the absence of this rigid
mechanical support, the bacterial cell literally explodes due to its high
internal osmotic pressure. he cell wall polysaccharide is made up of
alternating sugars, N-acetylglucosamine (NAG) and N-acetylmuramate
(NAM), linked together by glycosidic bonds (Figure 3–25). Lysozyme
normally cleaves after NAM units in the chain (that is, between NAM and
48
Chapter 3: Proteins
NAG
NAM
NAG
NAM
CH2OH
CH2OH
CH2OH
CH2OH
O
O
O
OH
N
O
O
O
OR
H
N
O
C
O
O
OH
H
N
C
O
Figure 3–25 Arrangement of NAM
and NAG in the bacterial cell-wall
polysaccharide (Problem 3–88).
OR
H
N
C
O
H
C
COO–
R= H
H2O
CH2OH
CH2OH
O
O
O
OH
CH2OH
C
CH3
CH2OH
O
OH
OR
O
O
OH
OR
HO
N
O
H
C
CH3
N
O
H
C
CH3
N
O
H
C
CH3
N
O
H
C
CH3
NAG), but will also cleave artiicial substrates composed entirely of NAG
units. When the crystal structure of lysozyme bound to a chain of three
NAG units (tri-NAG) was solved, it was discovered that the binding cleft
in lysozyme included six sugar-binding sites, A through F, and that triNAG illed the irst three of these sites. From the crystal structure it was
not apparent, however, which of the ive bonds between the six sugars
was the one that was normally cleaved. Tri-NAG is not cleaved by lysozyme, although longer NAG polymers are. It was clear from modeling
studies that NAM is too large to it into site C. Where are the catalytic
groups responsible for cleavage located relative to the six sugar-binding
sites?
A. Between sites A and B
B. Between sites B and C
C. Between sites C and D
D. Between sites D and E
E. Between sites E and F
Aspartate transcarbamoylase (ATCase) is an allosteric enzyme with six
catalytic and six regulatory subunits. It exists in two conformations:
one with low enzymatic activity and the other with high activity. In the
absence of any ligands, the low-activity conformation predominates.
Malate is an inhibitor of ATCase that binds in the active site at the position where the substrate aspartate normally binds. A very peculiar efect
of malate is observed when the activity of ATCase is measured at low
aspartate concentrations: there is an increase in ATCase activity at very
low malate concentrations, but then the activity decreases at higher concentrations (Figure 3–26).
A. How is it that malate, a bona ide inhibitor, can increase ATCase activity
under these conditions?
B. Would you expect malate to have the same peculiar efect if the measurements were made in the presence of a high concentration of aspartate?
Why or why not?
200
activity (percent)
3–89
100
0
0
5
10
malate (mM)
15
Figure 3–26 The activity of ATCase
with increasing concentrations of
the inhibitor malate (Problem 3–89).
These measurements were made at an
aspartate concentration well below its Km.
PROTEIN FUNCTION
49
TABLE 3–3 The observed dissociation constants (Kd values) of Cdk2 for ATP,
ADP, cyclin A, and histone H1 (Problem 3–90).
Kd (μM)
ATP
ADP
Cyclin A
Histone H1
Cdk2
0.25
1.4
0.05
not detected
P-Cdk2
0.12
6.7
0.05
–
+
+
–
–
+
–
–
–
+
cyclin A
–
–
+
+
+
histone H1
+
+
+
+
+
1
2
3
4
5
histone H1
100
Cdk2 + cyclin A
1.0
P-Cdk2 + cyclin A
0.7
3–90
Cyclin-dependent protein kinase 2 (Cdk2) regulates critical events in the
progression of the cell cycle in mammalian cells. Cdk2 can form a complex with cyclin A and can be phosphorylated by another protein kinase
to produce P-Cdk2. To determine the roles of cyclin A and phosphorylation in the function of Cdk2, you purify nonphosphorylated and phosphorylated Cdk2. You mix these two forms of Cdk2 and cyclin A in various combinations with 32P-ATP and assay for phosphorylation of histone
H1 (Figure 3–27). You also measure the binding ainity of various forms
of Cdk2 for ATP, ADP, cyclin A, and histone H1 (Table 3–3).
A. From Figure 3–27, what is required for Cdk2 to phosphorylate histone H1
eiciently?
B. How do the requirements identiied in part A speciically afect the function of Cdk2 relative to its target, histone H1 (Table 3–3 and Figure 3–27)?
C. he usual intracellular concentrations of ATP and ADP are in the range
0.1 to 1 mM. Assume that the binding of cyclin A to Cdk2 or P-Cdk2 does
not alter the ainities of either form of Cdk2 for ATP and ADP. Is it likely
that the observed changes in ainity for ATP and ADP are important for
Cdk2 function? Why or why not?
3–91
β-Catenin is a target for phosphorylation by glycogen synthase kinase
3 and also a substrate for ubiquitylation, which triggers degradation in
proteasomes. Treatment of mouse ibroblasts with a proteasome inhibitor, ALLN, increases the stability of β-catenin and causes the appearance
of new, slower migrating forms of the protein on sodium dodecyl sulfate
(SDS) polyacrylamide gels (Figure 3–28, lanes 1 and 2). Do these new
bands represent phosphorylated proteins, which often run more slowly
on such gels, or do they arise by addition of ubiquitin, which would
increase their size? To test for phosphorylation, you treat samples with
a protein phosphatase that eiciently removes phosphates from proteins and run them on gels (lanes 3 and 4). To test for ubiquitylation,
you express in cells a modiied form of ubiquitin that carries six histidine residues at its C-terminus, allowing its easy puriication via Ni2+column chromatography. You treat these cells with ALLN (or not), and
then purify His-ubiquitylated (His-Ub) proteins before running them on
gels (lanes 5 and 6). In all cases, you detect β-catenin speciically, using
antibodies directed against it.
Are the slower migrating forms of β-catenin due to phosphorylation
or ubiquitylation? Explain your answer.
3–92
Cdk2
P-Cdk2
Genome sequencing has revealed a surprisingly large number of protein
tyrosine phosphatases (PTPs), very few of which have known roles in the
life of a cell. PTP1B was the very irst member of the PTP family to be discovered, and its three-dimensional structure and catalytic mechanism
are well deined. PTP1B will dephosphorylate almost any phosphotyrosine-containing protein or peptide in the test tube, but in the cell it is
Figure 3–27 Phosphorylation of histone
H1 by various combinations of Cdk2
and cyclin A (Problem 3–90). The amount
of radioactive phosphate attached to
histone H1 in lanes 1 and 3 is 0.3% and
0.2%, respectively, of that in lane 5.
His-Ub
phosphatase
ALLN
+
+
+
+
+
+
+
1
2
3
4
5
6
190
108
89
Figure 3–28 Electrophoretic analysis
of β-catenin (Problem 3–91). β-Catenin
was detected using β-catenin-speciic
antibodies. Size markers in kilodaltons
are shown on the left.
50
Chapter 3: Proteins
102
2244
Tyr46 → Leu
4160
1700
155
Glu115 → Ala
5700
45
212
Lys120 → Ala
19,000
80
708
Asp181 → Ala
0.61
126
0.023
20
26
His214 → Ala
700
Cys215 → Ser
no activity
Arg221 → Lys
11
80
0.41
Arg221 → Met
3.3
1060
0.12
likely to have better-deined targets. One strategy for identifying a binding target is to incubate the known protein with a cell lysate, and then
isolate the known protein and identify any proteins that are bound to it.
An enzyme such as PTP1B, however, binds and releases its target as part
of its catalytic cycle. You reason that if you interfere with catalysis you
might be able to increase the dwell time of the substrate, making it stable
enough for isolation. To this end, you make several mutants of PTP1B
by changing speciic amino acids in the active site, and measure their
kinetic parameters (Table 3–4).
You pick mutant Cys215 → Ser (C215S) for study because the –SH
group of Cys215 initiates a nucleophilic attack on the phosphotyrosine,
releasing the phosphate. You pick a second mutant—mutant 2—because
it has promising kinetic properties. You use genetic engineering tricks to
fuse these two mutants and the wild-type enzyme to glutathione S-transferase (GST), so that each of the three proteins can be rapidly puriied by
precipitation with glutathione-Sepharose. You express the GST-tagged
proteins, precipitate them from cell lysates, separate the precipitated
proteins by gel electrophoresis, and identify phosphotyrosine-containing proteins using anti-phosphotyrosine antibodies. As shown in Figure
3–29, GST-mutant 2 bound two phosphotyrosine-containing proteins
(the two dark bands in lane 3), whereas GST-C215S bound none.
A. Which mutant PTP1B in Table 3–4 is likely to correspond to GST-mutant
2? Why do you think this protein gave a successful result?
B. Why might GST-C215S have failed to precipitate any phosphotyrosinecontaining proteins?
MEDICAL LINKS
3–93
he Ras protein is a GTPase that functions in many growth-factor signaling pathways. In its active form, with GTP bound, it transmits a downstream signal that leads to cell proliferation; in its inactive form, with
GDP bound, the signal is not transmitted. Mutations in the gene for Ras
are found in many cancers. Of the choices below, which alteration of Ras
activity is most likely to contribute to the uncontrolled growth of cancer
cells?
A. A mutation that prevents Ras from being made.
B. A mutation that increases the ainity of Ras for GDP.
vector control
60,200
Wild type
GST-M2
kcat (min–1)
Vmax
[nmol/(min/mg)]
GST-C215S
Km (nM)
Enzyme
GST-PTP1B
TABLE 3–4 Kinetic parameters of purified PTP1B mutants (Problem 3–92).
1
2
3
4
kd
200
116
97
87
Figure 3–29 Phosphotyrosine-containing
proteins precipitated with GST-tagged
PTP1B enzymes (Problem 3–92). GSTPTP1B is the fused wild-type protein;
GST-M2 is the fused mutant 2 protein. The
positions of “marker” proteins of known
molecular masses are indicated on the left.
PROTEIN FUNCTION
C. A mutation that decreases the ainity of Ras for GTP.
D. A mutation that decreases the ainity of Ras for its downstream targets.
E. A mutation that decreases the rate of hydrolysis of GTP by Ras.
3–94
he activity of Ras is carefully regulated by two other proteins, a guanine
nucleotide exchange factor (GEF) that stimulates the uptake of GTP by
Ras, and a GTPase-activating protein (GAP) that stimulates the hydrolysis of GTP by Ras. he activities of these regulatory proteins are in turn
also regulated. Which of the following changes in GAP and GEF proteins
might cause a cell to proliferate excessively?
A. A nonfunctional GAP
B. A permanently active GAP
C. A nonfunctional GEF
D. A permanently active GEF
MCAT STYLE
Passage 1 (Questions 3–95 to 3–99)
E2F protein binds speciic DNA sequences via its DNA-binding domain and activates transcription of genes that promote cell division. Another protein called
Rb binds to E2F and inhibits it, thereby preventing cell division. When the cell
receives a signal to initiate cell division, a protein kinase called Cdk binds Rb and
phosphorylates it. Cdk binds to Rb via a binding site that is separate from the
kinase catalytic site. Phosphorylation of Rb causes the Rb–E2F complex to dissociate, releasing E2F to activate transcription of cell-division genes. Rb is thus an
inhibitor of cell division. Inactivation of Rb by mutation or deletion is an important event in the genesis of many kinds of cancer.
3–95
Which of the following best describes the deining features of a protein
domain?
A. A discrete structural and functional unit within a protein
B. A group of proteins that share related functions
C. An evolutionarily conserved sequence of amino acids
D. he region of the cell in which a protein functions
3–96
Imagine a mutation that inactivates the site on Cdk that binds to Rb.
What would be the efect of this mutation on the reaction in which Cdk
phosphorylates Rb?
A. A large decrease in Km
B. A large increase in Km
C. A large decrease in Vmax
D. A large increase in Vmax
3–97
Which one of the following amino acids can be phosphorylated?
A. Alanine
B. Leucine
C. Lysine
D. Serine
3–98
Phosphorylation can afect the structure or function of a protein in which
of the following ways?
I. Charge attraction between the phosphate group and positively charged
amino acids causes a conformational change in the protein.
II. Charge repulsion between the phosphate group and negatively charged
amino acids causes a conformational change in the protein.
III. he phosphate group creates a binding site for another protein.
A. I
B. III
C. I and II
D. I, II, and III
51
Chapter 3: Proteins
52
3–99
Which of the following hypotheses provides the most likely explanation
for how phosphorylation of Rb causes its dissociation from E2F?
A. Phosphorylation of Rb causes E2F to change conformation so that it dissociates from Rb.
B. Phosphorylation of Rb causes it to undergo a conformational change that
blocks the E2F binding site.
C. Phosphorylation of Rb inactivates the Rb binding site for Cdk.
D. Phosphorylation of Rb leads to a charge repulsion between the phosphate group and a positively charged pocket on E2F.
Passage 2 (Questions 3–100 to 3–103)
Steroid hormones such as estrogen, testosterone, and cortisol are small, difusible signaling molecules that control diverse aspects of cell physiology. Steroid
hormones signal to cells via receptor proteins. Typically, a steroid hormone diffuses across the cell membrane and binds to a receptor protein in the cytoplasm.
Binding of the steroid hormone to the receptor protein causes the receptor to bind
to transcription factor proteins that control transcription. In this manner, steroid
hormones activate or repress genes that control cellular physiology.
3–100
A.
B.
C.
D.
Which of the following best describes how a steroid hormone might
cause the receptor to bind to proteins that control transcription?
he steroid hormone acts catalytically to promote the binding reaction
between the receptor and the transcription factor.
he steroid hormone decreases the Km of the binding reaction between
the receptor and the transcription factor.
he steroid hormone induces the receptor to undergo a conformational
change that exposes a binding site for the transcription factor.
he steroid hormone promotes positive feedback to ensure that more
and more receptor binds to the transcription factor.
What kinds of interaction are likely to be responsible for binding of the
receptor protein to the transcription factor protein?
I. Hydrophobic forces
II. Noncovalent bonds
III. Covalent bonds
3–101
A. I
B. III
C. I and II
D. I, II, and III
All of the following are correct statements about common features of
binding interactions between proteins EXCEPT:
A. At equilibrium, proteins often undergo frequent association and dissociation.
B. Binding reactions usually require an input of energy from ATP hydrolysis.
C. Some proteins, once they have bound one another, remain tightly associated.
D. he rate of binding does not usually depend on the strength of the interaction.
3–102
Which one of the following approaches would provide information about
the strength of the interaction between the steroid hormone and its
receptor?
A. Measure the equilibrium constant
B. Measure the rate of association
C. Measure the rate of dissociation
D. Measure the Vmax for the binding reaction
3–103
Chapter 4
53
CHAPTER
DNA, Chromosomes,
and Genomes
4
THE STRUCTURE AND FUNCTION OF DNA
IN THIS CHAPTER
TERMS TO LEARN
antiparallel
base pair
complementary
deoxyribonucleic acid (DNA)
double helix
gene
genome
template
DEFINITIONS
Match each deinition below with its term from the list above.
4–1
4–2
he totality of the genetic information carried in the DNA of a cell or an
organism.
he three-dimensional structure of DNA, in which two DNA chains held
together by hydrogen bonds between the bases are coiled around one
another.
4–3
Information-containing element that controls a discrete hereditary characteristic.
4–4
Describes the relative orientation of the two strands in a DNA helix; the
polarity of one strand is oriented in the opposite direction to that of the
other.
4–5
Two nucleotides in an RNA or DNA molecule that are held together by
hydrogen bonds.
THE STRUCTURE AND
FUNCTION OF DNA
CHROMOSOMAL DNA AND ITS
PACKAGING IN THE CHROMATIN
FIBER
CHROMATIN STRUCTURE AND
FUNCTION
THE GLOBAL STRUCTURE OF
CHROMOSOMES
HOW GENOMES EVOLVE
H
N
H
N
N
TRUE/FALSE
4–8
4–9
N
N
N
O
adenine
thymine
Human cells do not contain any circular DNA molecules.
H
THOUGHT PROBLEMS
4–7
H
N
Decide whether this statement is true or false, and then explain why.
4–6
CH3
O
he start of the coding region for the human β-globin gene reads
5ʹ-ATGGTGCAC-3ʹ. What is the sequence of the complementary strand
for this segment of DNA?
Upon returning from a recent trip abroad, you explain to the customs
agent that you are bringing in a sample of DNA, deoxyribonucleic acid.
He is aghast that you want to bring an acid into his country. What is the
acid in DNA? Should the customs agent be wary?
he chemical structures for an A-T and a G-C base pair are shown in Figure 4–1, along with their points of attachment to the sugar-phosphate
backbones.
O
H
N
N
H
N
N
N
N
N
N
H
O
H
guanine
cytosine
Figure 4–1 An A-T and a G-C base pair,
as viewed along the helix axis (Problem
4–9). Each base is attached to its
deoxyribose sugar via the line extending
from the ring nitrogen.
54
Chapter 4: DNA, Chromosomes, and Genomes
A. Indicate the positions of the major and minor grooves of the DNA helix
on these representations.
B. Draw the structure of a T-A base pair in the same way as in Figure 4–1.
C. Do the same chemical moieties (for example, the methyl group of thymine) always project into the same groove?
4–10
O
Examine the space-illing models of the base pairs shown in Figure 4–2.
Each base pair includes two bases, two deoxyribose sugars, and two
phosphates. Can you identify the locations of the purine base, the pyrimidine base, the sugars, and the phosphates? Identify each base pair as
C-G, G-C, T-A, or A-T.
(A)
A
CH2 O
O
–O P O
O
C
CH2 O
(B)
O
–O P O
O
T
CH2 O
Figure 4–2 Space-illing models of two base pairs (Problem 4–10). Carbon and
phosphorus atoms are gray and orange, respectively, nitrogen atoms are blue, and
oxygen atoms are red. No hydrogen atoms are shown.
4–11
DNA isolated from the bacterial virus M13 contains 25% A, 33% T, 22% C,
and 20% G. Do these results strike you as peculiar? Why or why not? How
might you explain these values?
4–12
A segment of DNA from the interior of a single strand is shown in
Figure 4–3. What is the polarity of this DNA from top to bottom?
4–13
DNA forms a right-handed helix. Pick out the right-handed helix from
those shown in Figure 4–4.
O
Figure 4–3 Three nucleotides from
the interior of a single strand of DNA
(Problem 4–12). Arrows at the ends of the
DNA strand indicate that the structure
continues in both directions.
CALCULATIONS
4–14
Human DNA contains 20% C on a molar basis. What are the mole percents of A, G, and T?
4–15
he diploid human genome comprises 6.4 × 109 bp and its into a nucleus
that is 6 μm in diameter.
A. If base pairs occur at intervals of 0.34 nm along the DNA helix, what is the
length of DNA in a human cell?
B. If the diameter of the DNA helix is 2.4 nm, what fraction of the volume
of the nucleus is occupied by DNA? [Volume of a sphere is (4/3)πr3 and
volume of a cylinder is πr2h.]
4–16
(A)
(B)
(C)
One gram of cultured human cells contains about 109 cells and occupies
roughly 1 mL. If the average molecular mass of a base pair is 660 daltons
and each cell contains 6.4 × 109 bp, what mass of DNA is present in this
one-gram sample? If all the DNA molecules in the sample were laid endto-end to form a single thread, do you suppose it would be long enough
to reach from the Earth to the Moon (385,000 kilometers)?
DATA HANDLING
4–17
Bacteriophage T4 attaches to its bacterial host and injects its DNA to initiate an infection that ultimately releases hundreds of progeny virus. In
1952, Alfred Hershey and Martha Chase radiolabeled the DNA of bacteriophage T4 with 32PO43– and the proteins with 35S-methionine. hey
then mixed the labeled bacteriophage with bacteria and after a brief time
agitated the mixture vigorously in a blender to detach T4 from the bacteria. hey then separated the phage from the bacteria by centrifugation.
hey demonstrated that the bacteria contained 30% of the 32P label but
Figure 4–4 Three “DNA” helices
(Problem 4–13).
CHROMOSOMAL DNA AND ITS PACKAGING IN THE CHROMATIN FIBER
virtually none of the 35S label. When new bacteriophage were released
from these bacteria, they were also found to contain 32P but no 35S. How
does this experiment demonstrate that DNA rather than protein is the
genetic material? (Note that bacteriophage T4 contains only protein and
DNA.)
CHROMOSOMAL DNA AND ITS PACKAGING IN THE
CHROMATIN FIBER
TERMS TO LEARN
cell cycle
centromere
chromatin
chromosome
exon
histone
histone H1
homologous chromosome
(homolog)
intron
karyotype
nucleosome
replication origin
telomere
DEFINITIONS
Match each deinition below with its term from the list above.
4–18
Full set of chromosomes of a cell arranged with respect to size, shape,
and number.
4–19
Constricted region of a mitotic chromosome that holds sister chromatids
together.
4–20
Any one of a group of small abundant proteins, rich in arginine and
lysine, that form the primary level of chromatin organization.
4–21
Structure composed of a very long DNA molecule and associated proteins that carries part (or all) of the hereditary information of an organism.
4–22
he orderly sequence of events by which a cell duplicates its contents
and divides into two.
4–23
Complex of DNA, histones, and non-histone proteins found in the
nucleus of a eukaryotic cell.
4–24
One of the two copies of a particular chromosome in a diploid cell, each
copy being derived from a diferent parent.
4–25
Beadlike structure in eukaryotic chromatin, composed of a short length
of DNA wrapped around a core of histone proteins.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
4–26
Human females have 23 diferent chromosomes, whereas human males
have 24.
4–27
In the living cell, chromatin usually adopts the extended “beads-on-astring” form.
4–28
he four core histones are relatively small proteins with a very high proportion of positively charged amino acids; the positive charge helps the
histones bind tightly to DNA, regardless of its nucleotide sequence.
4–29
Nucleosomes bind DNA so tightly that they cannot move from the positions where they are irst assembled.
THOUGHT PROBLEMS
4–30
In the 1950s, the techniques for isolating DNA from cells all yielded molecules of about 10,000 to 20,000 base pairs. We now know that the DNA
55
56
Chapter 4: DNA, Chromosomes, and Genomes
molecules in all cells are very much longer. Why do you suppose such
short pieces were originally isolated?
4–31
Consider the following statement: a human cell contains 46 molecules of
DNA in its nucleus. Do you agree with it? Why or why not?
4–32
Chromosome 3 in orangutans difers from chromosome 3 in humans by
two inversion events that occurred in the human lineage (Figure 4–5).
Draw the intermediate chromosome that resulted from the irst inversion
and explicitly indicate the segments included in each inversion.
4–33
Deine a “gene.”
4–34
List the three specialized DNA sequences and their functions that act to
ensure that the number and morphology of chromosomes are constant
from one generation of a cell to the next.
4–35
Histone proteins are among the most highly conserved proteins in
eukaryotes. Histone H4 proteins from a pea and a cow, for example, differ in only 2 of 102 amino acids. However, comparison of the two gene
sequences shows many more diferences. hese observations indicate
that mutations that change amino acids must be selected against. Why
do you suppose that most amino acid-altering mutations in histone
genes are deleterious?
4–36
Duplex DNA composed entirely of CTG/CAG trinucleotide repeats (5ʹCTG in one strand and 5ʹ-CAG in the other strand) is unusually lexible. If
75 CTG/CAG repeats are incorporated into a much longer DNA molecule
and mixed with histone octamers, the irst nucleosome that assembles
nearly always includes the CTG/CAG repeat region. Can you suggest a
reason why CTG/CAG repeats might be such efective elements for positioning nucleosomes? (Consider what the energy of the binding interaction between the histone octamers and the DNA must accomplish.)
CALCULATIONS
4–37
At mitosis, human chromosome 1 is condensed to a form that measures
only 10 μm in length.
A. Given that this chromosome contains 2.8 × 108 bp, how compacted is the
DNA molecule in chromosome 1 at mitosis relative to its fully extended
length as naked DNA? (Recall that a DNA base pair is 0.34 nm in length.)
B. A single nucleosome core particle is 11 nm long and contains 147 bp
of DNA (0.34 nm/bp). What packing ratio (DNA length to nucleosome length) has been achieved by wrapping DNA around the histone
octamer? What fraction of the condensation that occurs at mitosis does
this irst level of packing represent?
C. Assuming that the 30-nm chromatin iber contains about 20 nucleosomes (200 bp/nucleosome) per 50 nm of length, calculate the degree
of compaction of DNA associated with this level of chromatin structure.
What fraction of the condensation that occurs at mitosis does this level of
DNA packing represent?
4–38
he total number of protein-coding genes in the human genome can
be calculated in several ways. It is important to remember that all such
numbers are estimates at present, because it is diicult to identify a gene
just from the DNA sequence. Chromosome 22 has about 700 genes in
48 Mb of sequence, which represents 1.5% of the 3200 Mb in the haploid genome. Using these numbers, how many genes would you estimate
for the haploid human genome? If your estimate is signiicantly larger or
smaller than the accepted value of approximately 21,000 protein-coding
genes, suggest possible explanations for the discrepancy.
two inversions
orangutan
human
Figure 4–5 Chromosome 3 in orangutans
and humans (Problem 4–32). Differently
colored blocks indicate segments of the
chromosomes that are homologous in
DNA sequence.
CHROMOSOMAL DNA AND ITS PACKAGING IN THE CHROMATIN FIBER
57
he 700 genes on chromosome 22 (48 Mb) average 19,000 bp in length
and contain an average of 5.4 exons, each of which averages 266 bp. On
average, what fraction of each gene sequence is converted to mRNA?
What fraction of the whole chromosome do genes occupy?
4–39
2500 kb
950 kb
Assuming that the histone octamer forms a cylinder 9 nm in diameter
and 5 nm in height and that the human genome forms 32 million nucleosomes, what volume of the nucleus (6 μm in diameter) is occupied by
histone octamers? [Volume of a cylinder is πr2h; volume of a sphere is
(4/3)πr3.] What fraction of the nuclear volume do the DNA and the histone octamers occupy (see Problem 4–15)?
4–40
610 kb
220 kb
DATA HANDLING
One way to demonstrate that a chromosome contains a single DNA molecule is to use a technique called pulsed-ield gel electrophoresis, which
can separate DNA molecules up to 107 bp in length. Ordinary gel electrophoresis cannot separate such long molecules because the steady electric ield stretches them out so they travel end-irst through the gel matrix
at a rate that is independent of their length. If the electric ield is changed
periodically, however, the DNA molecules are forced to re-orient to the
new ield before continuing their snakelike movement through the gel.
he time for re-orientation depends on length, so that longer molecules
move more slowly through the gel.
he results of pulsed-ield gel electrophoresis of the DNA from the
yeast Saccharomyces cerevisiae are shown in Figure 4–6. How many chromosomes does S. cerevisiae have?
4–42
A classic experiment attached telomeres from Tetrahymena to a linearized yeast plasmid, allowing the plasmid to grow as a linear molecule—
that is, as an artiicial chromosome (Figure 4–7). A circular 9-kb plasmid
was constructed to contain a yeast origin of replication (Ars1) and the
yeast Leu2 gene. Cells that are missing the chromosomal Leu2 gene, but
have taken up the plasmid, can grow in medium lacking leucine. he
plasmid was linearized with BglII, which cuts once (Figure 4–7), and then
mixed with 1.5-kb Tetrahymena telomere fragments generated by cleavage with BamHI (Figure 4–7). he mixture was incubated with DNA ligase
and the two restriction nucleases, BglII and BamHI. he ligation products included molecules of 10.5 kb and 12 kb in addition to the original
components. he 12-kb band was puriied and transformed into yeast,
which were then selected for growth in the absence of leucine. Samples
of DNA from one transformant were digested with HpaI, PvuII, or PvuI
and the fragments were separated by gel electrophoresis and visualized
after hybridization to a plasmid-speciic probe (Figure 4–8).
Figure 4–6 Pulsed-ield gel electrophoresis
of S. cerevisiae chromosomes (Problem
4–41). To minimize the handling of DNA,
which would surely break it, the cells
themselves are placed at the top of
the gel and gently opened by addition
of a lysis buffer. The DNA molecules in
this gel have been exposed to the dye
ethidium bromide, which luoresces under
ultraviolet light when it is bound to DNA.
This treatment allows the DNA—otherwise
invisible—to be seen.
BamHI
ke
I
uI
kb
telomere
Leu2
Ars1
HpaI
PvuII
telomere
PvuI
PvuI
HpaI
m
ar
s1
Pv
I
pa
H
Ar
L
Pv
BglII
2
eu
uI
rs
4–41
9.4
6.5
4.4
PvuII
9-kb plasmid
with yeast gene
Tetrahymena
telomere fragment
2.3
2.0
intended linear chromosome
Figure 4–7 Structure of 9-kb plasmid, telomere fragment, and the intended linear chromosome
with Tetrahymena telomeres (Problem 4–42). The sites of unique cutting by restriction enzymes
are indicated.
Figure 4–8 Autoradiograph of restriction
analysis of plasmid structure (Problem
4–42). Marker DNAs of known sizes are
shown on the right.
58
Chapter 4: DNA, Chromosomes, and Genomes
A. How do the results of the analysis in Figure 4–8 distinguish between the
intended linear form of the plasmid in the transformed yeast and the
more standard circular form?
B. Explain how ligation of the DNA fragments in the presence of the restriction nucleases BamHI and BglII ensures that you get predominantly the
linear construct you want. he recognition site for BglII is 5ʹ-A*GATCT-3ʹ
and for BamHI is 5ʹ-G*GATCC-3ʹ, where the asterisk (*) is the site of cutting.
4–43
Nucleosomes can be assembled onto deined DNA segments. When a
particular 225-bp segment of human DNA was used to assemble nucleosomes and then incubated with micrococcal nuclease, which digests
DNA that is not located within the nucleosome, uniform fragments
147 bp in length were generated. Subsequent digestion of these fragments with a restriction enzyme that cuts once within the original 225-bp
sequence produced two well-deined bands at 37 bp and 110 bp. Why do
you suppose two well-deined fragments were generated by restriction
digestion, rather than a range of fragments of diferent sizes? How would
you interpret this result?
4–44
You have been sent the irst samples of a newly discovered Martian microorganism for analysis of its chromatin. he cells resemble Earthly eukaryotes and are composed of similar molecules, including DNA, which is
located within a nucleus-like structure in the cell. One member of your
team has identiied two basic histone-like proteins associated with the
DNA in roughly an equal mass ratio with the DNA. You isolate nuclei from
the cells and treat them with micrococcal nuclease for various times. You
then extract the DNA and run it on an agarose gel alongside DNA from
rat-liver nuclei that had been briely digested with micrococcal nuclease.
As shown in Figure 4–9, the digest of rat-liver nuclei gives a standard ladder of nucleosomes, but the digest of the Martian organism gives a smear
of products with a nuclease-resistant limit of about 300 nucleotides. As a
control, you isolate the Martian DNA free of all protein and digest it with
micrococcal nuclease: it is completely susceptible, giving predominantly
mono- and dinucleotides.
Do these results suggest that the Martian organism has nucleosomelike structures in its chromatin? If so, what can you deduce about their
spacing along the DNA?
4–45
Moving nucleosomes out of the way is important for turning genes on. In
yeast the SWI/SNF complex, which is the founding member of the ATPdependent chromatin remodeling complexes, is required for both activating and repressing gene transcription. How does it work? In principle,
it could slide nucleosomes along the chromosome, knock them of the
DNA, or transfer them from one duplex to another.
To investigate this problem, you assemble a nucleosome on a 189-bp
segment of labeled DNA, which you then ligate to a longer piece of DNA
tethered to a magnetic bead, as shown in Figure 4–10A. You incubate
this substrate with the SWI/SNF complex either before or after you cut
the DNA with NheI, which cleaves near the nucleosome, or with EcoRI,
which cleaves near the bead. You use a magnet to separate the DNA that
is still attached to the bead from the released DNA. If the nucleosome is
present, the released DNA will run with a slower mobility on an agarose
gel than if it is absent. Incubation with the SWI/SNF complex in the presence of ATP, followed by NheI digestion (NheI 2nd), releases most of the
label as the free DNA fragment (Figure 4–10B, lane 6). By contrast, incubation with SWI/SNF after cleavage with NheI (NheI 1st) releases most of
the label as nucleosome-bound DNA (lane 4). Similar experiments with
EcoRI digestion showed that incubation with SWI/SNF released most of
the label as nucleosome-bound DNA regardless of whether EcoRI cleavage preceded or followed the incubation.
digestion time (minutes)
0.5
1
2
4
8 15 30
rat
base
pairs
1400
1200
1000
800
600
400
200
Figure 4–9 Micrococcal digest of
chromatin from a Martian organism
(Problem 4–44). The results of digestion of
rat-liver chromatin are shown on the right.
CHROMATIN STRUCTURE AND FUNCTION
59
(B)
(A)
magnetic
bead
nucleosome
32PO
4
EcoRI
NheI
NheI (1st)
– + – + – –
ATP
– + + + – +
SWI/SNF
NheI (2nd)
– – – + + +
– – + – + +
nucleosomebound DNA
Figure 4–10 The action of the SWI/SNF
complex on a nucleosome (Problem
4–45). (A) Nucleosome-containing
substrate tethered to a magnetic bead.
Sites of cleavage by the restriction
enzymes NheI and EcoRI are indicated.
(B) Results of incubation with the SWI/
SNF complex before (NheI 2nd) or after
(NheI 1st) cleavage with NheI. In the
absence of cleavage (lane 1), the labeled
substrate does not enter the gel.
free DNA
1 2 3 4 5 6
Do your results distinguish among the three possible mechanisms for
SWI/SNF action—nucleosome sliding, release, or transfer? Explain how
your results argue for or against each of these mechanisms.
MEDICAL LINKS
An abnormal human karyotype is shown in Figure 4–11. his particular
karyotype is found in the cancer cells of more than 90% of patients with
chronic myelogenous leukemia. Arrows indicate two abnormal chromosomes. Describe the event that led to this abnormal karyotype. Is this
patient male or female?
4–46
CHROMATIN STRUCTURE AND FUNCTION
TERMS TO LEARN
epigenetic inheritance
euchromatin
heterochromatin
position effect variegation
DEFINITIONS
Match each deinition below with its term from the list above.
4–47
Less condensed region of an interphase chromosome that stains diffusely.
4–48
Form of transmission of information from cell to cell, or from parent to
progeny, that is not encoded in DNA.
4–49
Diference in gene expression that depends on the location of the gene
on the chromosome.
2
1
3
6
7
8
13
14
15
19
20
9
21
22
4
5
10
11
12
16
17
18
X
Figure 4–11 Two karyotypes illustrating the typical alteration seen in chronic myelogenous leukemia (Problem 4–46).
60
Chapter 4: DNA, Chromosomes, and Genomes
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
4–50
Deacetylation of histone tails allows nucleosomes to pack together into
tighter arrays, which usually reduces gene expression.
4–51
Histone variants are often inserted into already formed chromatin.
THOUGHT PROBLEMS
4–52
Phosphorylation of serines and methylation and acetylation of lysines in
histone tails afect the stability of chromatin structure above the nucleosome level and have important consequences for gene expression. Draw
the structures for serine and phosphoserine, as well as those for lysine,
monomethylated lysine, and acetylated lysine. Which modiications alter
the net charge on a histone tail? Would you expect the changes in charge
to increase or decrease the ability of the tails to interact with DNA?
4–53
In contrast to histone acetylation, which always correlates with gene activation, histone methylation can lead to either transcriptional activation
or repression. How do you suppose that the same modiication—methylation—can mediate diferent biological outcomes?
4–54
Why is a chromosome with two centromeres (a dicentric chromosome)
said to be unstable? Wouldn’t a back-up centromere be a good thing for
a chromosome, giving it two chances to form a kinetochore and attach to
microtubules during mitosis? Wouldn’t that help to ensure that the chromosome didn’t get left behind at mitosis—sort of like using a belt and
braces to keep your pants up?
DATA HANDLING
4–55
he irst paper to demonstrate diferent chromatin structures in active
and inactive genes used nucleases to probe the globin loci in chicken
red blood cells, which express globin mRNA, and in chicken ibroblasts,
which do not. Isolated nuclei from these cells were treated with either
micrococcal nuclease or DNase I, and then DNA was prepared and
hybridized in vast excess to a 3H-thymidine-labeled globin cDNA. (A
cDNA is a DNA molecule made as a copy of mRNA.) If the nuclear DNA
has not been degraded, it will hybridize to the globin cDNA and protect
it from digestion by S1 nuclease, which is speciic for single strands of
DNA.
Digestion of red-cell nuclei or ibroblast nuclei with micrococcal
nuclease (so that about 50% of the DNA was degraded) yielded DNA samples that still protected greater than 90% of the cDNA from subsequent
digestion with S1 nuclease. Similarly, digestion of ibroblast nuclei with
DNase I (so that less than 20% was degraded) yielded DNA that protected
greater than 90% of the cDNA. An identical digestion of red-cell nuclei
with DNase I, however, yielded DNA that protected only about 25% of the
cDNA. hese results are summarized in Table 4–1.
When nucleosome monomers were isolated from red blood cells by
digestion with micrococcal nuclease, their DNA protected more than
90% of globin cDNA. If the monomers were irst treated with DNase I,
the isolated DNA protected only 25% of globin cDNA. If the monomers
were briely treated with trypsin to remove 20–30 amino acids from the
N-terminus of each histone, digestion of the modiied nucleosomes with
micrococcal nuclease yielded DNA that protected only 25% of globin
cDNA (Table 4–1).
A. Which nuclease—micrococcal nuclease or DNase I—digests chromatin
that is being expressed (active chromatin)? How can you tell?
CHROMATIN STRUCTURE AND FUNCTION
61
H3
H4
unmod K9-Me S10-P unmod
TABLE 4–1 Protection of globin cDNA by untreated and nuclease-treated
chromatin samples (Problem 4–55).
I UB I UB I UB I UB
Protected globin cDNA after treatment
Source of DNA
none
micrococcal
DNase I
Fibroblast nuclei
93%
91%
91%
Red-cell nuclei
95%
92%
25%
Red-cell nucleosomes
91%
25%
Red-cell nucleosomes (trypsin)
25%
B. Does trypsin treatment of nucleosome monomers appear to render a
random population or a speciic population of nucleosomes sensitive to
micrococcal nuclease? How can you tell?
C. Is the alteration that distinguishes active chromatin from bulk chromatin
a property of individual nucleosomes, or is it related to the way nucleosome monomers are packaged into higher-order structures within the
cell nucleus?
4–56
HP1 proteins, a family of proteins found in heterochromatin, are implicated in gene silencing and chromatin structure. he three proteins in
humans—HP1α, HP1β, and HP1γ—share a highly conserved domain
that directs their localization to chromatin. To determine whether these
proteins could bind to the histone H3 N-terminus, you have covalently
attached to separate beads various versions of the H3 N-terminal peptide—unmodiied, Lys9-dimethylated (K9-Me), and Ser10-phosphorylated (S10-P)—along with an unmodiied tail from histone H4. his
arrangement allows you to incubate the beads with various proteins,
wash away unbound proteins, and then elute bound proteins for assay
by Western blotting. he results of your “pull-down” assay for the HP1
proteins are shown in Figure 4–12, along with the results from several
control proteins, including Pax5, which is a gene regulatory protein,
polycomb protein Pc1, which is known to bind to histones, and Suv39h1,
a histone methyl transferase.
Based on these results, which of the proteins tested bind to the
unmodiied tails of histones? Do any of the HP1 proteins and control
proteins selectively bind to any of the various histone N-terminal peptides? What histone modiication would you predict would be found in
heterochromatin?
4–57
Look at the two yeast colonies in Figure 4–13. Each of these colonies contains about 100,000 cells descended from a single yeast cell, originally
telomere
Pax5
Pc1
Suv39h1
HP1α
HP1β
HP1γ
Figure 4–12 Pull-down assays to
determine binding speciicity of HP1
proteins (Problem 4–56). Each protein at
the left was detected by immunoblotting
using a speciic antibody after
separation by SDS polyacrylamidegel electrophoresis. For each histone
N-terminal peptide, the total input protein
(I), the unbound protein (U), and the
bound protein (B) are indicated.
telomere
Ade2 gene at normal location
white colony of
yeast cells
Ade2 gene moved near telomere
red colony of
yeast cells
with white sectors
Figure 4–13 Position effect on expression
of the yeast Ade2 gene (Problem 4–57).
The Ade2 gene codes for one of the
enzymes of adenosine biosynthesis,
and the absence of the Ade2 gene
product leads to the accumulation of a
red pigment. Therefore a colony of cells
that express Ade2 is white, and one
composed of cells in which the Ade2
gene is not expressed is red.
62
Chapter 4: DNA, Chromosomes, and Genomes
somewhere in the middle of the clump. A white colony arises when the
Ade2 gene is expressed from its normal chromosomal location. When the
Ade2 gene is moved to a location near a telomere, it is packed into heterochromatin and inactivated in most cells, giving rise to colonies that are
mostly red. In these largely red colonies, white sectors fan out from the
middle of the colony. In both the red and white sectors, the Ade2 gene is
still located near telomeres. Explain why white sectors have formed near
the rim of the red colony. Based on the patterns observed, what can you
conclude about the propagation of the transcriptional state of the Ade2
gene from mother to daughter cells in this experiment?
4–58
High-density DNA microarrays can be used to analyze changes in expression of all the genes in the yeast genome in response to various perturbations. he efects of depletion of histone H4 and deletion of the Sir3
gene on expression of all yeast genes were analyzed in this way, as summarized on the yeast chromosomes illustrated in Figure 4–14. Depletion
of histone H4 was achieved in a strain in which the gene was engineered
to respond to galactose. In the absence of galactose, the gene is turned
of and depletion of histone H4 is evident within 6 hours, leading to a
decreased density of nucleosomes throughout the genome. Deletion of
Sir3 removes a critical component of the Sir protein complex that binds
to telomeres and is responsible for deacetylation of telomeric nucleosomes.
A. Depletion of histone H4 signiicantly increased expression of 15% of all
yeast genes (Figure 4–14A, black bars). Does loss of histone H4 increase
expression of a greater fraction of genes near telomeres than in the rest of
the genome? Explain how you arrived at your conclusion.
(A)
HISTONE H4 DEPLETION
I
II
III
IV
V
VI
VII
VIII
IX
X
XI
XII
XIII
XIV
XV
XVI
(B)
Sir3 DELETION
I
II
III
IV
V
VI
VII
VIII
IX
X
XI
XII
XIII
XIV
XV
XVI
Figure 4–14 Changes in expression of the genes on chromosomes I–XVI of yeast (Problem 4–58). (A) In response to depletion of
histone H4. (B) In response to deletion of Sir3. Black bars indicate genes whose expression was increased relative to wild-type yeast
threefold or more. The very light gray bars show genes whose expression was decreased by threefold or more; they are not relevant
to this problem but have been added for the sake of completeness. Chromosomes were split at their centromeres so that all their
telomeres could be aligned at the left; brackets indicate the pairs of arms that make up individual chromosomes. Three chromosome
arms have been shortened to it into the igure, as indicated by diagonal lines.
CHROMATIN STRUCTURE AND FUNCTION
63
om
os
N
3
D
d
id
na
ke
2
m
as
pl
m
id
A
m
ro
1
as
pl
ch
id
e
m
as
tiv
pl
A classic paper examined the arrangement of nucleosomes around the
centromere (CEN3) of yeast chromosome III. Because centromeres
are the chromosome attachment sites for microtubules, it was unclear
whether they would have the usual arrangement of nucleosomes. his
study used plasmids into which were cloned various lengths of the native
chromosomal DNA around the centromere (Figure 4–15). Chromatin
from native yeasts and from yeasts that carried individual plasmids was
treated briely with micrococcal nuclease, and then the DNA was deproteinized and digested with the restriction enzyme BamHI, which cuts the
DNA only once (Figure 4–15). he digested DNA was fractionated by gel
electrophoresis and analyzed by Southern blotting using a segment of
radiolabeled centromeric DNA as a hybridization probe (Figure 4–15).
his procedure (called indirect end labeling) allows visualization of all
the DNA fragments that include the DNA immediately to the right of the
BamHI cleavage site. As a control, a sample of naked DNA from the same
region was treated with micrococcal nuclease and subjected to the same
analysis. An autoradiogram of the results is shown in Figure 4–16.
A. When the digestion with BamHI was omitted, a regular, though much
less distinct, set of dark bands was apparent. Why does digestion with
BamHI make the pattern so much clearer and easier to interpret?
B. Draw a diagram showing the micrococcal-nuclease-sensitive sites on
the chromosomal DNA and the arrangement of nucleosomes along the
chromosome. What is special about the centromeric region?
C. What is the purpose of including a naked DNA control in the experiment?
D. he autoradiogram in Figure 4–16 shows that the native chromosomal
DNA yields a regularly spaced pattern of bands beyond the centromere;
that is, the bands at 600 nucleotides and above are spaced at 160-nucleotide intervals. Does this regularity result from the lining up of nucleosomes at the centromere, like cars at a stoplight? Or, is the regularity an
intrinsic property of the DNA sequence itself? Explain how the results
with plasmids 1, 2, and 3 decide the issue.
na
4–59
e
B. Deletion of the Sir3 gene signiicantly increased expression of 1.5% of all
yeast genes (Figure Figure 4–14B, black bars). Does the absence of Sir3
protein increase expression of a greater fraction of genes near telomeres
than in the rest of the genome? Explain how you arrived at your conclusion.
C. If you concluded that either the depletion of histone H4 or the deletion of
the Sir3 gene, or both, preferentially increased expression of genes near
telomeres, propose a mechanism for how that might happen.
base
pairs
1560
1400
1240
1080
920
760
600
BamHI
native
chromosome
CEN3
plasmid 1
350
bacterial DNA
plasmid 2
yeast DNA
plasmid 3
probe
190
1 kb
Figure 4–15 Diagrams of the structures of the native chromosome and three plasmids
(Problem 4–59). The native chromosome is linear; its true ends extend well beyond the
positions marked by the diagonal lines. The plasmids, which are circular, are shown here
as linear molecules for ease of comparison. The native yeast sequences around the
centromere are shown as thin lines. Bacterial DNA sequences in the plasmids are shown
as red rectangles. The yeast DNA in plasmid 3, which is shown as a pink rectangle, is a
segment of yeast chromosomal DNA far removed from the centromere.
Figure 4–16 Results of micrococcalnuclease digestion of DNA around CEN3
(Problem 4–59). Approximate lengths of
DNA fragments in nucleotide pairs are
indicated on the left of the autoradiogram.
64
Chapter 4: DNA, Chromosomes, and Genomes
THE GLOBAL STRUCTURE OF CHROMOSOMES
TERMS TO LEARN
lampbrush chromosome
mitotic chromosome
nucleolus
polytene chromosome
DEFINITIONS
Match each deinition below with its term from the list above.
4–60
Giant chromosome in which the DNA has undergone repeated replication without separation into new chromosomes.
4–61
Paired chromosomes in meiosis in immature amphibian eggs, in which
the chromatin forms large stif loops extending out from the linear axis of
the chromosome.
4–62
Highly condensed, duplicated chromosome with the two new chromosomes still held together at the centromere as sister chromatids.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
4–63
In lampbrush chromosomes of amphibian oocytes, most of the DNA is
in the loops, which are actively transcribed, while the rest remains highly
condensed on the chromosome axis, where genes are generally not
expressed.
4–64
At the inal level of condensation each chromatid of a mitotic chromosome is organized into loops of chromatin that emanate from a central
axis.
THOUGHT PROBLEMS
4–65
Although mammalian chromosomes have not yet been found to form
lampbrush chromosomes in amphibian oocytes, chromosomes from different amphibians form typical lampbrush chromosomes when injected
into oocytes in the form of demembranated sperm heads. When sperm
heads from Rana pipiens (Northern leopard frog), which forms large
loops in its own oocyte chromosomes, were injected in Xenopus laevis
oocytes, the resulting lampbrush chromosomes had the small loops typical of those in X. laevis oocytes. Similarly, when sperm heads from X.
laevis were injected into Notophthalmus viridescens (red spotted newt)
oocytes, the resulting lampbrush chromosomes had the very large loop
structure typical of N. viridescens.
Do these heterologous injection experiments support the idea that
loop structure is an intrinsic property of a chromosome? Why or why not?
DATA HANDLING
4–66
he characteristic banding patterns of the giant polytene chromosomes
of Drosophila melanogaster provide a visible map of the genome that
has proven an invaluable aid in genetic studies for decades. he molecular basis for the more intense staining of bands relative to interbands
remains a puzzle. In principle, bands might stain more darkly because
they contain more DNA than interbands due to overreplication, or the
amount of DNA may be the same in the bands and interbands, but the
DNA stains more prominently in the bands because it is more condensed
THE GLOBAL STRUCTURE OF CHROMOSOMES
65
Figure 4–17 Autoradiograph of blot-hybridization analysis of polytene
and diploid chromosomes (Problem 4–66). P and D refer to polytene and
diploid, respectively. Numbers at the top refer to cloned DNA segments
used as probes: 2851 and 2842 are from the 315-kb region under analysis
(see Figure 4–18); 2148 is from elsewhere in the genome and was used in
all hybridizations to calibrate the amount of DNA added to the gels.
or contains more proteins. hese two possibilities—diferential replication or diferential staining—were distinguished by the experiments
described below.
A series of radiolabeled segments spanning 315 kb of a Drosophila
chromosome were used as hybridization probes to estimate the amount
of corresponding DNA present in normal diploid tissues versus DNA
from salivary glands, which contain polytene chromosomes. DNA samples from diploid and polytene chromosomes were digested with combinations of restriction enzymes. he fragments were then separated by gel
electrophoresis and transferred to nitrocellulose ilters for hybridization
analysis. In every case the restriction pattern was the same for the DNA
from diploid chromosomes and polytene chromosomes, as illustrated
for two examples in Figure 4–17. he intensities of many speciic restriction fragments were measured and expressed as the ratio of the intensity
of the fragment from polytene chromosomes to the intensity of the corresponding fragment from diploid chromosomes (Figure 4–18).
How do these results distinguish between diferential replication and
diferential staining as the basis for the diference between bands and
interbands? Explain your reasoning.
4–67
he typical coiled phone cord provides an everyday example of the phenomenon of supercoiling. Invariably, the cord becomes coiled about
itself forming a tangled mess. hese coiled coils are supercoils. Dangling
the receiver and letting it spin until it stops can remove them. Similarly,
supercoils can be reintroduced by twisting the receiver, which of course
is how they get there in the irst place.
DNA is coiled into a double helix that exhibits the same phenomenon
of supercoiling (Figure 4–19). A relaxed circular DNA, with 10.5 bp per
turn of the helix, will assume a more-or-less circular form when laid onto
2.0
relative
hybridization 1.0
intensity
0
0
50
100
150
200
250
300 kilobases
2842
2851
5
87 D
11 12
14
1,2
87 E
3
4
5,6
band
number
Figure 4–18 Relative amounts of DNA in diploid and polytene chromosomes at
different points along the chromosome (Problem 4–66). The chromosomal segment
covered by the cloned restriction fragments is shown at the bottom, along with
the cytological designations for the chromosome regions and bands. The cloned
fragments are shown above the chromosome, and the positions of 2851 and 2842
are indicated. The ratio of hybridization of each restriction fragment to DNA from
polytene chromosomes versus diploid chromosomes is plotted above each fragment.
2851
P D
2148
P D
2842
P D
2148
P D
66
Chapter 4: DNA, Chromosomes, and Genomes
negative
supercoils
relaxed
circle
positive
supercoils
1
2
3
4
5
relaxed
–2
or
solenoidal
(left-handed)
plectonemic
(right-handed)
+2
or
solenoidal
plectonemic
(left-handed) (right-handed)
Figure 4–19 Relaxed and supercoiled circular DNA molecules (Problem 4–67).
Duplexes of DNA are indicated by single lines. These DNA molecules differ only in
the number of times one strand is wound around the other, a quantity known as the
linking number. Solenoidal supercoils are shown as wrapped around a cylinder for
illustrative purposes.
a surface. If one strand of the DNA is broken and wound around its partner two extra times (overwound—an increase in linking number of +2)
and then rejoined, the molecule will twist on itself, forming two positive
supercoils. If one strand in a relaxed circular DNA is broken and rejoined
with two fewer turns (underwound—a decrease in linking number of
–2), the molecule will twist to form two negative supercoils. Positive and
negative supercoils each can assume two forms termed plectonemic and
solenoidal, although plectonemic supercoils are the only ones that are
stable in naked DNA. he efect of supercoiling is to preserve the preferred local winding of DNA at 10.5 bp per turn. In cells, the degree of
supercoiling of DNA is carefully controlled by special enzymes called
topoisomerases that break and rejoin strands of DNA.
Circular plasmid DNA isolated from E. coli is highly supercoiled, as is
evident when the DNA is separated by electrophoresis on an agarose gel
(Figure 4–20, lane 1). When incubated for increasing times with E. coli
topoisomerase I (which breaks and reseals a single DNA strand in negatively supercoiled DNA but not in positively supercoiled DNA), several
new bands appear between the supercoiled DNA and the relaxed DNA
(lanes 2–5).
A. In the untreated DNA sample isolated from bacteria, why do you suppose a small fraction of the plasmid molecules are relaxed (Figure 4–20,
lane 1)?
B. What are the discrete bands between the highly supercoiled and relaxed
bands in Figure 4–20 that appear with increasing times of incubation
with topoisomerase I? Why do they move at rates intermediate between
relaxed and highly supercoiled DNA?
C. Estimate the number of supercoils that were present in the original plasmid molecules.
D. Did the bacterial plasmid originally contain positive or negative supercoils? Explain your answer.
4–68
Imagine that you assemble a single nucleosome on a closed circular,
relaxed DNA molecule; that is, a circular duplex DNA with no breaks in
either strand and zero supercoiling. Wrapping the DNA molecule around
the histone octamer forms solenoidal supercoils, which are compensated for by plectonemic supercoils in another part of the molecule.
Of the four possible arrangements of solenoidal and plectonemic
supercoils shown in Figure 4–21, which have a net supercoiling of zero?
(Since no breaks were introduced into the DNA in the process of forming
the nucleosome, it must retain an overall supercoiling of zero.) Indicate
the sign of the supercoiling (positive or negative) on the structures you
select.
supercoiled
Figure 4–20 Plasmid DNA treated with
E. coli topoisomerase I for increasing
times (Problem 4–67). Plasmid DNA is
shown before incubation with E. coli
topoisomerase I in lane 1 and after
increasing times of incubation in lanes 2–5.
(A)
(B)
(C)
(D)
Figure 4–21 Four possible arrangements
of circular DNA molecules with two
solenoidal and two plectonemic supercoils
(Problem 4–68). The position of the
nucleosome is indicated by the cylinder.
HOW GENOMES EVOLVE
4–69
67
Which of the two alternative arrangements of compensating solenoidal
and plectonemic supercoils, generated by the formation of a nucleosome (see Problem 4–68), represents the true biological situation? hese
alternatives were distinguished by incubating the nucleosome-bound
DNA with either E. coli topoisomerase I, which can remove only negative plectonemic supercoils, or with calf thymus topoisomerase I, which
can remove both negative and positive plectonemic supercoils. Histones
were removed after the incubation with a topoisomerase, and the presence of supercoils in the naked DNA was assayed by gel electrophoresis
(see Figure 4–20, Problem 4–67). (he sign of the plectonemic supercoils
in the naked DNA can be determined by subsequent incubation with E.
coli topoisomerase I, which relaxes negative supercoils but not positive
ones.)
It was found that incubation of nucleosomal DNA with E. coli topoisomerase I gave DNA molecules with zero supercoils. By contrast, incubation with calf thymus topoisomerase I gave DNA molecules with two
negative supercoils. Are the solenoidal supercoils around biological
nucleosomes positive (right-handed) or negative (left-handed)? What
results would you have expected for the other alternative?
HOW GENOMES EVOLVE
TERMS TO LEARN
copy number variation (CNV)
homologous
pseudogene
purifying selection
single-nucleotide polymorphism (SNP)
DEFINITIONS
Match each deinition below with its term from the list above.
4–70
A copy of a functional gene that has become irreversibly inactivated by
multiple mutations.
4–71
Long blocks of DNA sequence that difer in the number of times they are
present in the genomes of diferent individuals in a population.
4–72
Evolutionary process that eliminates individuals carrying mutations that
interfere with important genetic functions.
4–73
Variation between individuals at a certain nucleotide position in the
genome.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
4–74
In a comparison between the DNAs of related organisms such as humans
and mice, identifying the conserved DNA sequences facilitates the search
for functionally important regions.
4–75
Many human genes so closely resemble their homologs in yeast that the
protein-coding portion of the human gene will substitute for the function
of the yeast gene in yeast cells.
4–76
he portion of the human genome subjected to purifying selection corresponds to the protein-coding sequences.
4–77
Gene duplication and divergence is thought to have played a critical role
in the evolution of increased biological complexity.
Chapter 4: DNA, Chromosomes, and Genomes
68
Figure 4–22 Transposable elements and
genes in 1-Mb regions of chromosomes
2 and 22 (Problem 4–78). Blue lines that
project upward indicate exons of known
genes. Red lines that project downward
indicate transposable elements; they are
so numerous that they merge into nearly
a solid block outside the Hox clusters.
chromosome 22
chromosome 2
100 kb
HoxD cluster
THOUGHT PROBLEMS
Mobile pieces of DNA—transposable elements—that insert themselves
into chromosomes and accumulate during evolution make up more than
40% of the human genome. Transposable elements are inserted moreor-less randomly throughout the human genome. hese elements are
conspicuously rare at the four homeobox gene clusters, HoxA, HoxB,
HoxC, and HoxD, as illustrated for HoxD in Figure 4–22, along with an
equivalent region of chromosome 22, which lacks a Hox cluster. Each
Hox cluster is about 100 kb in length and contains 9–11 genes, whose
diferential expression along the anteroposterior axis of the developing
embryo establishes the basic body plan for humans (and for other animals). Why do you suppose that transposable elements are so rare in the
Hox clusters?
4–78
CALCULATIONS
4–79
Nucleotide sequence comparisons are fundamental to our current conception of the tree of life, to our understanding of how mitochondria
and chloroplasts were acquired and their subsequent evolution, to the
importance and magnitude of horizontal gene transfer, and to the notion
that by focusing on a few model organisms we can gain valid insights into
all of biology. For these reasons we’ve designed this problem and the
following one to introduce the common methods and assumptions that
underlie the art of nucleotide sequence comparison.
A phylogenetic tree represents the history of divergence of species
from common ancestors. Construction of such trees from DNA or protein sequences can really only be done with computers: the data sets are
enormous and the algorithms are subtle. Nevertheless, some of the fundamental principles of tree construction can be illustrated with a simple
example. Consider the irst 30 amino acids of the hemoglobin α chains
for the ive species shown in Figure 4–23.
A. In a common approach, known as the distance-matrix method, the
irst step is to construct a table of all pairwise diferences between the
sequences. A partially illed-in example is shown in Table 4–2. Complete
the table by illing in the blanks indicated by question marks.
TABLE 4–2 The difference matrix for the first 30 amino acids of the hemoglobin
α chains from five species (Problem 4–79).
Human
Frog
Chicken
Whale
Fish
Human
VLSPADKTNVKAAWGKVGAHAGEYGAEALE
Frog
LLSADDKKHIKAIMPAIAAHGDKFGGEALY
Chicken VLSAADKNNVKGIFTKIAGHAEEYGAETLE
Human
Frog
Chicken
Whale
Fish
Whale
VLSPTDKSNVKATWAKIGNHGAEYGAEALE
0
?
11
8
17
Fish
SLSDKDKAAVRALWSKIGKSADAIGNDALS
0
?
17
20
0
?
20
0
?
0
Figure 4–23 Alignment of the irst
30 amino acids of the hemoglobin
α chains from ive species (Problem 4–79).
Amino acids are represented by the
one-letter code (see Table 8, page 966).
Amino acids that differ from the human
sequence are highlighted in yellow.
HOW GENOMES EVOLVE
Human
69
I
II
a
b
III
IV
V
d
c
f
g
e
Figure 4–24 A general phylogenetic
tree for ive species (Problem 4–79).
Figure 4–25 A general phylogenetic
tree for ive species with line segments
indicated by letters (Problem 4–80).
B. According to the information in the completed Table 4–2, which pair of
species is most closely related? What is the assumption that underlies
your choice?
C. he information in Table 4–2 can be used to arrange species on the phylogenetic tree shown in Figure 4–24. he branching order is determined
using a simple kind of cluster analysis. he two most similar species are
placed on the adjacent branches at the upper left in Figure 4–24. he species with the fewest average diferences relative to this pair is placed on
the next branch. In the next step, these three species are combined and
the average diferences from the remaining species are calculated and
used to ill in the next branch, and so on. Use this method to arrange the
species on the tree in Figure 4–24.
D. Is the branching order you determined in part C the same as you would
get by simply using the number of diferences relative to human to place
the other species on the tree? Why is the method of cluster analysis superior?
4–80
In the previous question the branching order, or topology, of the tree
was established, but actual distances (number of diferences) were not
assigned to the line segments that make up the tree (Figure 4–25). To
calculate distances for line segments is, once again, tedious by hand but
easy by computer. he following exercise gives a feeling for how such calculations are done.
A. Using the numbers from the completed distance matrix in Table 4–2 and
the branching order determined in the previous problem, write down all
the equations for the diferences between species in terms of the line segments that make up the tree (Figure 4–25). Are there enough equations to
solve for the lengths of the seven line segments?
B. One straightforward, not-too-exhausting method for solving these equations is to consider them three at a time; for example,
I → II = a + b
I → III = a + c + d
II → III = b + c + d
here are 10 such three-at-a-time equations for ive species. Using the
information in the distance-matrix table (see Table 4–2), solve two of
these sets of equations—human/whale/chicken and human/whale/
frog—for a and b. Are the values for a and b the same in the two solutions?
DATA HANDLING
4–81
he earliest graphical method for comparing nucleotide sequences—the
so-called diagon plot—still yields one of the best visual comparisons of
sequence relatedness. An example is illustrated in Figure 4–26, where
the human β-globin gene is compared with the human cDNA for β
70
Chapter 4: DNA, Chromosomes, and Genomes
(A) HUMAN β-GLOBIN cDNA
COMPARED WITH HUMAN
β-GLOBIN GENE
5′
5′
mouse β-globin gene
3′
human β-globin cDNA 3′
(B) MOUSE β-GLOBIN GENE
COMPARED WITH
HUMAN β-GLOBIN GENE
5′
human β-globin gene
3′
globin (Figure 4–26A) and with the mouse β-globin gene (Figure 4–26B).
(A cDNA is a DNA molecule made as a copy of mRNA and therefore lacking the introns that are present in genomic DNA.) Diagon plots are generated by comparing blocks of sequence, in this case blocks of 11 nucleotides at a time. If 9 or more of the nucleotides match, a dot is placed on
the diagram at the coordinates corresponding to the blocks being compared. A comparison of all possible blocks generates diagrams such as
the ones shown in Figure 4–26, in which sequence homologies show up
as diagonal lines.
A. From the comparison of the human β-globin gene with the human
β-globin cDNA (Figure 4–26A), deduce the positions of exons and introns
in the β-globin gene.
B. Are the entire exons of the human β-globin gene (indicated by shading
in Figure 4–26B) homologous to the mouse β-globin gene? Identify and
explain any discrepancies.
C. Is there any homology between the human and mouse β-globin genes
that lies outside the exons? If so, identify its location and ofer an explanation for its preservation during evolution.
D. Has either of the genes undergone a change of intron length during their
evolutionary divergence? How can you tell?
4–82
Your irst foray into archaeological DNA studies ended in embarrassment. he dinosaur DNA sequences that you so proudly announced to
the world later proved to be derived from contaminating modern human
cells—probably your own. Setting your sights slightly lower, you decide
to try to amplify residual mitochondrial DNA from a well-preserved
Neanderthal skeleton. You also redesign your laboratory to minimize
the possibility of stray contamination. You carefully prepare three diferent samples (A, B, and C) of bone from the femur and perform separate
polymerase chain reactions (PCR) on them, one in the laboratory of a
foreign collaborator. Sure enough, clear products of the expected size are
seen in all three reactions. Cloned products from each PCR are individually sequenced, with the results shown in Figure 4–27. he sequence of
the corresponding region of mitochondrial DNA from a human is shown
at the top. Dots indicate matches to the human sequence; dashes indicate missing DNA.
To determine whether the common sequence diferences you observe
could be due to normal variation within the human population, you make
pairwise comparisons of your consensus (most common) Neanderthal
sequence with a large number of individual human sequences. You
do the same for individual human sequences versus one another and
versus chimpanzee sequences. Your pairwise comparisons are shown in
Figure 4–28.
A. Have you successfully identiied a Neanderthal mitochondrial DNA
sequence? Explain your reasoning.
5′ human β-globin gene 3′
Figure 4–26 Diagon plots (Problem 4–81).
(A) Human β-globin cDNA compared with
the human β-globin gene. The β-globin
cDNA is a complementary DNA copy of
the β-globin mRNA. (B) Mouse β-globin
gene compared with the human β-globin
gene. The positions of the exons in the
human β-globin gene are indicated by
shading in (B). The 5ʹ and 3ʹ ends of the
sequences are indicated. The human
gene sequence is identical in the two
plots. To accommodate the short β-globin
cDNA sequence (549 nucleotides) and
the sequence of the β-globin gene
(2052 nucleotides) in similar spaces,
while maintaining proportional scales
within each plot, the scale of (A) is about
three times that of (B).
HOW GENOMES EVOLVE
71
Human
A1
A2
A3
A4
A5
A6
A7
A8
A9
A10
A11
A12
A13
A14
A15
A16
A17
A18
ACAGCAATCAACCCTCAACTATCACACATCAACTGCAACTCCAAAGCCACCCCT-CACCCAC
.............T......-...T.........A...........A.GTT.T.A.......
.............T......G...T.........A...........A.G...T.G.......
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.T.......A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
........T....T......G...T......G..A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.T.......A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T....T.G...T.T.......A...........A.G...T.A.......
.............T..........T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T..........T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
............................................G.........-.......
B1
B2
B3
B4
B5
B6
B7
B8
B9
B10
B11
B12
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.T...........T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.......T.AT..........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T....T.G...T.........A...........A.G...T.A..T....
.............T......G...T.........A.....T.....A.G...T.A.......
......................................................-.......
......................................................-.......
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
C13
C14
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A.......
.T......T....T......G...T...C.....A...........A.G...T.A.......
.T......T....T......G...T.........A...........A.G...T.A..T....
.T......T....T......G...T.........A...........A.G...T.A..T....
.............T......G...T.........A...........A.G...T.A..T....
.T.....CT....T......G...T.........A...........A.G...T.A.......
.T......T....T......G...T.........A...........A.G...T.A.......
.............T......G...T.........A...........A.G...T.A..T....
.......C..............................................-.......
......................................................-.......
......................................................-.......
.............T........................................-.......
......................................................-.......
Figure 4–27 Sequences of mitochondrial
DNA derived from Neanderthal samples
(Problem 4–82). The sequence across the
top is a reference human sequence. Dots
indicate matches to the human sequence;
dashes indicate missing nucleotides.
B. Did your extensive precautions in handling the sample eliminate human
contamination?
C. What is the reason for choosing mitochondrial DNA for archaeological
DNA studies? Wouldn’t nuclear DNA sequences be more informative?
D. What would you consider to be the most important way to conirm or
refute your indings?
Alu sequences are present at six sites in the introns of the human serum
albumin and α-fetoprotein genes. (hese genes are evolutionary relatives
25
human
human
human
Neanderthal
human
chimpanzee
20
percent of pairs
4–83
15
10
5
0
0
10
20
30
40
number of differences
50
60
Figure 4–28 Pairwise comparisons of
DNA sequences (Problem 4–82). The
human–human distribution compared
994 individual, and distinct, human
sequences with one another. These 994
sequences represent contemporary
human mitochondrial lineages; that is,
distinct sequences occurring in one or
more individuals. The fraction of pairs
with a given number of differences
is plotted. The human–Neanderthal
comparison is one Neanderthal sequence
against the 994 contemporary human
sequences. The human–chimpanzee
comparison involved 986 contemporary
human sequences with 16 contemporary
chimpanzee sequences.
72
Chapter 4: DNA, Chromosomes, and Genomes
Alu repeat
(300 nucleotides)
TTAAATAGGCCGGG----------AAAAAAAAAAAAATTAAATA
TGTGTGGGGATCAGG----------AAAAAAAAAAAAATCTGTGGG
TCTTCTTAGGCTGGG----------GAAAAAAAAAAAATCTTCTTA
ATAATAGTATCTGTCGGCTGGG----------AGAAAAAAAAAAATAAATAGTATCTGTC
GGATGTTGTGGGGCCGGG----------AAAAAAAAAAAAAGGATGTTGTGG
AGAACTAAAAGGGCTAGG----------AAAAAAGAGAAGAAGAACCGAAAG
Figure 4–29 Nucleotide sequences of the six Alu inserts in the human albumin-gene
family (Problem 4–83). Dashed lines indicate nucleotides in the internal part of the
Alu sequences.
that are located side by side in mammalian genomes.) he same pair
of genes in the rat contain no Alu sequences. he lineages of rats and
humans diverged more than 85 million years ago at the time of the mammalian radiation. Does the presence of Alu sequences in the human
genes and their absence from the corresponding rat genes mean that Alu
sequences invaded the human genes only recently, or does it mean that
the Alu sequences have been removed in some way from the rat genes?
To examine this question, you have sequenced all six of the Alu
sequences in the human albumin-gene family. he sequences around
the ends of the inserted Alu elements are shown in Figure 4–29.
A. Alu sequences create duplications of several nucleotides on each side of
the target site where they insert. Mark the left and right boundaries of
the inserted Alu sequences and indicate the nucleotides in the lanking
chromosomal DNA that have been altered by mutation.
B. he rate of nucleotide substitution in introns has been measured at about
3 × 10–3 mutations per million years at each site. Assuming the same rate
of substitution into the intron sequences that were duplicated by these
Alu sequences, calculate how long ago the Alu sequences inserted into
these genes. (Lump all the Alu sequences together to make this calculation; that is, treat them as if they inserted at about the same time.)
C. Why are these particular lanking sequences used in the calculation?
Why were larger segments of the intron not included? Why were the
mutations in the Alu sequences themselves not used?
D. Did these Alu sequences invade the human genes recently (after the time
of the mammalian radiation), or have they been removed from the rat
genes?
MEDICAL LINKS
4–84
About 5% of the human genome consists of duplicated segments of chromosomes, many of which are highly homologous, indicating a relatively
recent origin. he high degree of homology occasionally allows inappropriate recombination events to occur between the duplications, which
can decrease or increase the number of duplicated segments. Such
events are responsible for several human diseases, including the red–
green color blindness that afects 8% of the male population. he genes
for the red and green visual pigments lie near one another on the X chromosome, one in each copy of the duplicated segment. hey are 98% identical throughout most of their length, in both exons and introns; however,
the genes can be distinguished by the chance presence of an extra RsaI
cleavage site in one of the two genes. As a result, digestion with RsaI gives
a longer fragment for the RsaI-A gene than for the RsaI-B gene, and this
so-called restriction fragment length polymorphism (RFLP) can be used
to track the two genes (Figure 4–30).
A. To determine which gene encodes which pigment, several normal, redblind, and green-blind males were screened using a hybridization probe
normal
vision
individual 1
2
3
greenblind
4
5
6
7
redblind
8
9 10
RsaI-A
RsaI-B
Figure 4–30 RsaI RFlPs in normal, greenblind, and red-blind males (Problem 4–84).
RsaI-A refers to the RFlP characteristic
of the RsaI-A gene; RsaI-B refers to the
RFlP characteristic of the RsaI-B gene.
Individual males are indicated by number.
HOW GENOMES EVOLVE
9
73
individual
7
3
4
ratio A:B
0:1 1:0 1:1 2:1 3:1
1
2
4:1
kb
222
183
144
Figure 4–31 NotI digests of DNA
from selected normal and color-blind
individuals (Problem 4–84). Numbers for
individuals correspond to the numbers in
Figure 4–30. The ratios of RsaI-A genes
to RsaI-B genes (A:B) were estimated
from the intensity of hybridization in
Figure 4–30. The sizes of NotI fragments
are indicated in kb.
1
2
3
4
5
6
7
8
105
66
speciic for the RsaI RFLP (Figure 4–30). Which gene encodes the red
visual pigment, and which encodes the green visual pigment?
B. he intensity of hybridization in normal individuals was constant for the
RsaI-B gene, but surprisingly variable for the RsaI-A gene. his anomaly
was investigated by digesting the DNA from selected individuals with
NotI, which cleaves once within the RsaI-A gene but does not cleave the
RsaI-B gene. he restriction fragments were separated by pulsed-ield gel
electrophoresis and hybridized with a probe that recognizes both genes
(Figure 4–31). What is the basis for the variable intensity of hybridization
of RsaI-A genes in males with normal color vision? Can your explanation
account for the high frequency of color blindness?
C. What is the size of the duplicated chromosomal segment at this site in the
human genome?
4–85
Figure 4–32 DNA ingerprint analysis of
shufled twins (Problem 4–85).
here has been a colossal snafu in the maternity ward at your local hospital. Four sets of male twins, born within an hour of each other, were inadvertently shuled in the excitement occasioned by that unlikely event.
You have been called in to set things right. As a irst step, you want to get
the twins matched up. To that end you analyze a small blood sample from
each infant using a hybridization probe that detects polymorphic diferences in the numbers of simple sequence repeats such as (CA)n located
in widely scattered regions of the genome. he results are shown in
Figure 4–32.
A. Which infants are brothers?
B. How will you match brothers to the correct parents?
MCAT STYLE
Passage 1 (Questions 4–86 to 4–89)
Important advances in our understanding of epigenetics came from studies of
yeast mating. Haploid yeast cells exist in two mating types that are referred to
as a and α. Haploid cells of the opposite mating type can mate, which results in
fusion of the two cells to form a diploid cell. Two loci called HMLa and HMRα
contain the key information that controls mating behavior. he genes at these loci
are normally repressed, but they can be expressed when they are copied into the
MAT locus: if the information from HMLa is copied into the MAT locus, the cell
becomes an a cell, and if the information from HMRα is copied, the cell becomes
an α cell. his system allows cells to easily switch mating types.
4–86
Two short DNA sequences, called the E and I sites, which lank the
HML locus (Figure 4–33), control expression of the two genes at the
HML locus. Deletion of these sites causes the mating-type genes at the
E
I
a2
a1
tRNAThr
Figure 4–33 The HMl locus in yeast
(Problem 4–86). The HMl locus contains
two genes—a1 and a2—both of which
are copied into the MAT locus to change
the mating type.
74
Chapter 4: DNA, Chromosomes, and Genomes
HML locus to be expressed. If other, non-mating-type genes are placed
between the E and I sites, they become repressed. Which one of the following statements best explains the functions of the E and I sites?
A. hey are binding sites for chromatin remodeling complexes that inhibit
transcription.
B. hey are binding sites for factors that stimulate formation of heterochromatin.
C. hey are binding sites for histone chaperones that remove histones from
the HML locus.
D. hey are telomeric sequences that create a repressed chromatin structure.
4–87
Genetic analyses discovered that mutations in genes called SIR2, SIR3,
and SIR4 caused expression of genes at the HML locus. Biochemical
analysis showed that the SIR proteins form a complex. Which of the following activities might help to explain the roles of the SIR proteins in
repression of gene expression?
I. SIR2 is a histone deacetylase
II. SIR3 binds to deacetylated histones
III. SIR4 interacts with the E and I sites
A. I and II
B. I and III
C. II and III
D. I, II, and III
4–88
Which of the following could contribute to repression of gene expression
at the HML locus?
I. Binding of barrier proteins adjacent to HML
II. Formation of chromatin loops involving HML
III. Localization of HML near the nuclear envelope
A. I
B. I and II
C. II and III
D. III
4–89
A highly expressed tRNA gene is located just outside the HML. If this
gene is deleted, nearby genes that lie beyond the tRNA gene become
repressed. Which one of the following hypotheses could explain this
observation?
A. A site near the tRNA gene that is included in the deletion acts to position
the HML locus in the region near the nuclear envelope.
B. Nucleosomes at the tRNA gene recruit histone chaperones that block the
spread of repressive chromatin from the HML locus.
C. Mechanisms that maintain a low density of histones over the highly transcribed tRNA gene act as a barrier to the spread of repressive chromatin.
D. he tRNA gene is covered by heterochromatin that blocks spread of the
repressive chromatin from the HML locus.
Passage 2 (Questions 4–90 and 4–91)
Cancer cells proliferate without control, leading to formation of tumors. To investigate the nature of the changes that cause normal cells to become cancer cells,
scientists carried out a series of nuclear transfer experiments. hey irst removed
nuclei from mouse oocytes (egg cells), using a micropipette, and then replaced
the egg nucleus with a nucleus from one of several diferent kinds of mouse cancer cells. he reconstituted oocytes were put back into a mouse and allowed to
undergo early stages of development. hese experiments showed that nuclei from
several kinds of cancer cells were able to support early development of a mouse
embryo into multiple diferentiated cell types, without signs of abnormal proliferation.
HOW GENOMES EVOLVE
4–90
Which one of the following hypotheses best explains these experimental
results?
A. Epigenetic changes at centromeres prevent chromosome segregation,
which blocks proliferation of the cancer cells.
B. Epigenetic changes cause the cancer cell chromosomes to become highly
condensed, inhibiting expression of cancer-causing genes.
C. Epigenetic changes in gene expression that contribute to cancer are
erased when the nucleus is transferred to the mouse oocyte.
D. Mutations that cause cancer are repaired back to the wild-type sequence
when the nucleus is transferred to the mouse oocyte.
4–91
In another series of experiments, it was found that inactivation of both
copies of the p16 gene is a key step in the series of events that leads to
cancer. In many cancer cells, the p16 gene is inactivated by mutation or
deletion. However, in other cancer cells, the p16 gene contains no mutations or deletions, yet the p16 protein is not produced. Which of the following could explain the lack of p16 protein in these cancer cells?
I. Formation of euchromatin at the p16 gene
II. Formation of heterochromatin at the p16 gene
III. Modiication of histone tails at the p16 gene
A. I
B. II
C. I and III
D. II and III
75
The Chromosome of Escherichia coli.
John Cairns published this iconic
autoradiograph of the DNA of E. coli
caught in the act of replication in
1963. The bacteria were labeled with
3H-thymidine for “about two generations”
and then lysed with lysozyme in the
presence of carrier calf thymus DNA.
Most of the chromosomes detected after
two months’ exposure to ilm were “more
or less tangled” circles about 1.1 to
1.4 mm long. This one, by happy chance,
was unbroken and nicely displayed.
Cairns assumed, incorrectly as we now
know, that replication proceeded in
one direction from a ixed point, with a
“swivel” at the other y junction.
Chapter 5
77
DNA Replication, Repair,
and Recombination
THE MAINTENANCE OF DNA SEQUENCES
TERMS TO LEARN
germ cell
mutation
mutation rate
somatic cell
DEFINITIONS
Match each deinition below with its term from the list above.
5–1
A randomly produced, heritable change in the nucleotide sequence of a
chromosome.
5–2
Cell type in a diploid organism that carries only one set of chromosomes
and is specialized for sexual reproduction. A sperm or an egg.
5–3
Any cell of a plant or animal other than a germ cell or germ-line precursor.
TRUE/FALSE
Decide whether the statement is true or false, and then explain why.
5–4
Both germ-cell DNA stability and somatic-cell DNA stability are essential
for the survival of the species.
THOUGHT PROBLEMS
5–5
You infected an E. coli culture with a virulent bacterial virus (bacteriophage). Most of the cells lysed, but a few survived: 1 × 10–4 in your
sample. You wonder where the resistant bacteria came from. Were they
caused by the bacteriophage infection, or did they already exist in the bacterial culture? Earlier, for a diferent experiment, you had spread a dilute
suspension of E. coli onto solid medium in a large Petri dish, and, after
seeing that about 105 colonies were growing up, you made an imprint
of the colonies on that plate and transferred it to three other plates—a
process known as replica plating—which creates the same pattern of
colonies on the three plates. You realize that you can use these plates to
distinguish between the two possibilities. You pipette a suspension of the
bacteriophage onto each of the three replica plates so the bacteria can
be infected. What result do you expect if the bacteriophage cause resistance? What result do you expect if the resistant bacteria preexist?
5–6
he following statement sounds patently false. “he diferent cells in your
body rarely have genomes with the identical nucleotide sequence.” Provide an argument for why it might be true.
5–7
Individual organisms that carry harmful mutations tend to be eliminated
from a population by natural selection. It is easy to see how deleterious
CHAPTER
5
IN THIS CHAPTER
THE MAINTENANCE OF DNA
SEQUENCES
DNA REPLICATION MECHANISMS
THE INITIATION AND
COMPLETION OF
DNA REPLICATION IN
CHROMOSOMES
DNA REPAIR
HOMOLOGOUS RECOMBINATION
TRANSPOSITION AND
CONSERVATIVE SITE-SPECIFIC
RECOMBINATION
78
Chapter 5: DNA Replication, Repair, and Recombination
mutations in bacteria, which have a single copy of each gene, are eliminated by natural selection; the afected bacteria die and the mutation is
thereby lost from the population. Eukaryotes, however, have two copies of most genes because they are diploid. It is often the case that an
individual with two normal copies of the gene (homozygous, normal)
is indistinguishable in phenotype from an individual with one normal
copy and one defective copy of the gene (heterozygous). In such cases,
natural selection can operate only on an individual with two copies of the
defective gene (homozygous, defective). Imagine the situation in which a
defective form of the gene is lethal when homozygous, but without efect
when heterozygous. Can such a mutation ever be eliminated from the
population by natural selection? Why or why not?
CALCULATIONS
5–8
Mutations are introduced into the E. coli genome at the rate of about 1
mutation per 109 base pairs (bp) per generation. Imagine that you start
with a population of 106 E. coli, none of which carry any mutations in your
gene of interest, which is 1000 nucleotides in length and not essential for
bacterial growth and survival. In the next generation, after the population doubles in number, what fraction of the cells, on average, would you
expect to carry a mutation in your gene? After the population doubles
again, what would you expect the frequency of mutants in the population
to be? What would the frequency be after a third doubling?
DATA HANDLING
5–9
To determine the reproducibility of mutation frequency measurements,
you do the following experiment. You inoculate each of 10 cultures with
a single E. coli bacterium, allow the cultures to grow until each contains
106 cells, and then measure the number of cells in each culture that
carry a mutation in your gene of interest. You were so surprised by the
initial results that you repeated the experiment to conirm them. Both
sets of results display the same extreme variability, as shown in Table
5–1. Assuming that the rate of mutation is constant, why do you suppose
there is so much variation in the frequencies of mutant cells in diferent
cultures?
TABLE 5–1 Frequencies of mutant cells in multiple cultures (Problem 5–9).
Culture (mutant cells/106 cells)
Experiment
1
2
3
4
5
6
7
8
9
10
1
4
0
257
1
2
32
0
0
2
1
2
128
0
1
4
0
0
66
5
0
2
DNA REPLICATION MECHANISMS
TERMS TO LEARN
clamp loader
DNA helicase
DNA ligase
DNA polymerase
DNA primase
DNA topoisomerase
lagging strand
leading strand
replication fork
RNA primer
single-strand DNA-binding
(SSB) protein
sliding clamp
strand-directed mismatch
repair
DEFINITIONS
Match each deinition below to its term from the list above.
5–10
Short length of RNA synthesized on the lagging strand during DNA replication and subsequently removed.
DNA REPLICATION MECHANISMS
79
Enzyme that joins two adjacent DNA strands together.
5–11
5′
5–12
DNA repair process that replaces incorrect nucleotides inserted during
DNA replication.
5–13
Enzyme that opens the DNA helix by separating the single strands.
5–14
A protein complex that encircles the DNA double helix and binds to DNA
polymerase, keeping it irmly bound to the DNA while it is moving.
5–15
Enzyme that binds to DNA and reversibly breaks a phosphodiester bond
in one or both strands, allowing the DNA to rotate at that point.
5–16
Y-shaped region of a replicating DNA molecule at which the two daughter strands are formed.
5–17
he newly made strand of DNA found at a replication fork that is made in
discontinuous segments, which are later joined covalently.
3′
PO4– HO
Figure 5–1 A DNA fragment with a singlestranded gap on the bottom strand
(Problem 5–23).
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
5–18
When read in the same direction (5ʹ-to-3ʹ), the sequence of nucleotides
in a newly synthesized DNA strand is the same as in the parental strand
used as the template for its synthesis.
5–19
Each time the genome is replicated, half the newly synthesized DNA is
stitched together from Okazaki fragments.
5–20
In E. coli, where the replication fork travels at 500 nucleotide pairs per
second, the DNA ahead of the fork—in the absence of topoisomerase—
would have to rotate at nearly 3000 revolutions per minute.
5–21
Topoisomerase I does not require ATP to break and rejoin DNA strands
because the energy of the phosphodiester bond is stored transiently in a
phosphotyrosine linkage in the enzyme’s active site.
THOUGHT PROBLEMS
5–22
he nucleotide sequence of one DNA strand of a DNA double helix is
5ʹ-GGATTTTTGTCCACAATCA-3ʹ. What is the sequence of the complementary strand?
5–23
he DNA fragment in Figure 5–1 is double-stranded at each end but single-stranded in the middle. he polarity of the top strand is indicated. Is
the phosphate (PO4–) shown on the bottom strand at the 5ʹ end or the 3ʹ
end of the fragment to which it is attached?
5–24
Look carefully at the structures of the molecules in Figure 5–2.
A. What would you expect to happen if dideoxycytidine triphosphate
(ddCTP) were added to a DNA replication reaction in large excess over
(A) ddCTP
(B) ddCMP
NH2
NH2
N
N
O–
–
O
P
O–
O
P
O–
O
P
N
O
O–
O
–
O
CH2
P
N
O
O
O
O
O
O
O
H
H
O
CH2
H
H
Figure 5–2 Potential replication
substrates (Problem 5–24).
(A) Dideoxycytidine triphosphate (ddCTP).
(B) Dideoxycytidine monophosphate
(ddCMP).
80
Chapter 5: DNA Replication, Repair, and Recombination
the concentration of deoxycytidine triphosphate (dCTP)? Would it be
incorporated into the DNA? If it were, what would happen after that?
Give your reasoning.
B. What would happen if ddCTP were added at 10% of the concentration of
dCTP?
C. What efects would you expect if dideoxycytidine monophosphate
(ddCMP) were added to a DNA replication reaction in large excess, or at
10% of the concentration of dCTP?
5–25
How would you expect the loss of the 3ʹ-to-5ʹ proofreading exonuclease
activity of DNA polymerase in E. coli to afect the idelity of DNA synthesis? How would its loss afect the rate of DNA synthesis? Explain your
reasoning.
5–26
Discuss the following statement: “Primase is a sloppy enzyme that makes
many mistakes. Eventually, the RNA primers it makes are replaced with
DNA made by a polymerase with higher idelity. his is wasteful. It would
be more energy-eicient if a DNA polymerase made an accurate copy in
the irst place.”
5–27
SSB proteins bind to single-strand DNA at the replication fork and prevent the formation of short hairpin helices that would otherwise impede
DNA synthesis. What sorts of sequences in single-strand DNA might be
able to form a hairpin helix? Write out an example of a sequence that
could form a ive-nucleotide hairpin helix, and show the helix.
5–28
Conditional lethal mutations have proven indispensible in genetic and
biochemical analyses of complex processes such as DNA replication.
Temperature-sensitive (ts) mutations, which are one form of conditional
lethal mutation, allow cells to grow at one temperature (for example,
30°C) but not at a higher temperature (for example, 42°C).
A large number of temperature-sensitive replication mutants have
been isolated in E. coli. hese mutant bacteria are defective in DNA replication at 42°C but not at 30°C. If the temperature of the medium is raised
from 30°C to 42°C, these mutants stop making DNA in one of two characteristic ways. he “quick-stop” mutants halt DNA synthesis immediately,
whereas the “slow-stop” mutants stop DNA synthesis only after many
minutes.
A. Predict which of the following proteins, if temperature sensitive, would
display a quick-stop phenotype and which would display a slow-stop
phenotype. In each case, explain your prediction.
1. DNA topoisomerase I
2. A replication initiator protein
3. Single-strand DNA-binding protein
4. DNA helicase
5. DNA primase
6. DNA ligase
B. Cell-free extracts of the mutants show essentially the same patterns of
replication as the intact cells. Extracts from quick-stop mutants halt DNA
synthesis immediately at 42°C, whereas extracts from slow-stop mutants
do not stop DNA synthesis for several minutes after a shift to 42°C. Suppose extracts from a temperature-sensitive DNA helicase mutant and a
temperature-sensitive DNA ligase mutant were mixed together at 42°C.
Would you expect the mixture to exhibit a quick-stop phenotype, a slowstop phenotype, or a nonmutant phenotype?
5–29
DNA repair enzymes preferentially repair mismatched bases on the
newly synthesized DNA strand, using the old DNA strand as a template.
If mismatches were instead repaired without regard for which strand
served as template, would mismatch repair reduce replication errors?
Would such a mismatch repair system result in fewer mutations, more
DNA REPLICATION MECHANISMS
81
mutations, or the same number of mutations as there would have been
without any repair at all? Explain your answers.
5–30
If DNA polymerase requires a perfectly paired primer in order to add the
next nucleotide, how is it that any mismatched nucleotides “escape” this
requirement and become substrates for mismatch repair enzymes?
5–31
DNA damage can interfere with DNA replication. X-rays, for example,
generate highly reactive hydroxyl radicals that can break one or both
strands of DNA. Ultraviolet (UV) light commonly generates cyclobutane
dimers between adjacent T bases in the same DNA strand, which blocks
progression of DNA polymerase. If such damage is not repaired, it can
have serious consequences when a replication fork encounters it. See if
you can predict the appearance of the replication fork after it encounters
a nick or a thymine-dimer block in the templates for the leading and lagging strands. Replication forks just before they encounter the damage are
shown in Figure 5–3.
(B) UV LIGHT
(A) X-RAYS
5′
3′
I
5′
3′
III
5′
5′
nick
3′
3′
5′
5′
3′
3′
II
IV
5′
5′
3′
3′
5′
5′
nick
3′
5–32
thymine
dimer
3′
thymine
dimer
At the completion of replication of the circular genome of the animal
virus SV40, the two daughter circles are interlocked like links in a chain.
How do you suppose such interlinked molecules might then separate?
CALCULATIONS
5–33
Like all organisms, bacteriophage T4 encodes an SSB protein that is
important for removing secondary structure in the single-strand DNA
ahead of the replication fork. he T4 SSB protein is an elongated monomeric protein with a molecular weight of 35,000. It binds tightly to singlestrand, but not double-strand, DNA. Binding saturates at a 1:12 weight
ratio of DNA to protein. he binding of SSB protein to DNA shows a peculiar property that is illustrated in Figure 5–4. In the presence of excess
single-strand DNA (10 μg), virtually no binding is detectable at 0.5 μg
SSB protein (Figure 5–4A), whereas nearly all the SSB protein is bound to
DNA at 7.0 μg (Figure 5–4B).
A. At saturation, what is the ratio of nucleotides of single-strand DNA to
molecules of SSB protein? (he average mass of a single nucleotide is 330
daltons.)
Figure 5–3 Damage to the templates for
the leading and lagging strands (Problem
5–31). (A) X-ray-induced nicks. (B) UVlight-induced thymine dimers. Structures
I and III have damage in the template for
lagging-strand synthesis. Structures II
and IV have damage in the template for
leading-strand synthesis.
Chapter 5: DNA Replication, Repair, and Recombination
SSB (cpm)
SSB
protein
0.4
DNA
300
0.3
200
0.2
SSB
protein
100
0.1
0
0
top
Figure 5–4 Binding of T4 SSB protein
to single-strand DNA (Problem 5–33).
The binding of (A) 0.5 μg and (B) 7.0 μg
SSB protein to DNA was analyzed by
centrifugation through sucrose gradients,
on which the much more massive DNA
sediments more rapidly than protein
and is consequently found closer to the
bottom of the gradient (cpm stands for
counts per minute).
(B) 7.0 µg SSB
(A) 0.5 µg SSB
400
bottom
top
DNA (absorbance at 260 nm)
82
bottom
direction of sedimentation
B. When the binding of SSB protein to DNA reaches saturation, are adjacent
monomers of SSB protein likely to be in contact? Assume that a monomer of SSB protein extends for 12 nm along the DNA upon binding and
that the spacing of bases in single-strand DNA after binding is the same
as in double-strand DNA (that is, 10.4 nucleotides per 3.4 nm).
C. Why do you think that the binding of SSB protein to single-strand DNA
depends so strongly on the amount of SSB protein?
5–34
Approximately how many high-energy bonds are used to replicate the E.
coli chromosome? How many molecules of glucose would E. coli need to
consume to provide enough energy to copy its DNA once? How does this
mass of glucose compare with the mass of E. coli, which is about 10–12
g? (here are 4.6 × 106 base pairs in the E. coli genome. Oxidation of one
glucose molecule yields about 30 high-energy phosphate bonds. Glucose
has a molecular mass of 180 daltons, and there are 6 × 1023 daltons/g.)
DATA HANDLING
A born skeptic, you plan to conirm for yourself the results of a classic
experiment originally performed in the 1960s by Meselson and Stahl.
hey concluded that each daughter cell inherits only one strand of its
mother’s DNA. To check their results, you “synchronize” a culture of
growing cells, so that virtually all cells begin and then complete DNA synthesis at the same time. You irst grow the cells in a medium that contains
nutrients highly enriched in heavy isotopes of nitrogen and carbon (15N
and 13C in place of the naturally abundant 14N and 12C). Cells growing
in this “heavy” medium use the heavy isotopes to build all of their macromolecules, including nucleotides and nucleic acids. You then transfer
the cells to a normal, “light” medium containing 14N and 12C nutrients.
Finally, you isolate DNA from cells that have grown for diferent numbers
of generations in the light medium and determine the density of their
DNA by density-gradient centrifugation. Your data, plotting the amount
of DNA isolated versus its density, are shown in Figure 5–5. Are these
results in agreement with your expectations? Explain the results.
amount of DNA
5–35
starting cells
first generation
in light medium
light
light
second generation
third generation
6
4
2
0
heavy
heavy
light
heavy
light
heavy
Figure 5–5 Density of DNAs isolated from
cells that were grown for different times
in “light” medium after initial growth in
medium enriched for heavy isotopes
of nitrogen and carbon (Problem 5–35).
Equal culture volumes were analyzed
for each time point. Amount of DNA is
in arbitrary units, with the peak amount
of DNA in the sample containing starting
cells set equal to 1.
DNA REPLICATION MECHANISMS
5–36
5–37
83
he dnaB gene of E. coli encodes a helicase (dnaB) that unwinds DNA
at the replication fork. Its properties have been studied using artiicial
substrates like those shown in Figure 5–6. In such substrates, dnaB binds
preferentially to the longest single-strand region (the largest target) available. he experimental approach is to incubate the substrates under a
variety of conditions and then subject a sample to electrophoresis on
agarose gels. he short single strand (substrates 1 and 2) or strands (substrate 3) will move slowly if still annealed to the longer DNA strand, but
will move much faster if unwound and detached. he migration of these
short single strands can be followed selectively by making them radioactive and examining their positions in the gel by autoradiography. he
migration of the labeled single strands in the three diferent substrates is
shown in Figure 5–7. In the absence of any treatment, the labeled strands
move slowly (lanes 4, 8, and 12); when the substrates are heated to 100°C,
the labeled strands are detached and migrate more rapidly (lanes 3, 7,
and 11).
he results of several experiments are shown in Figure 5–7. Substrate
1, the substrate without tails, was not unwound by dnaB and ATP at 37°C
(Figure 5–7, lanes 1 and 2). When either substrate with tails was incubated at 37°C with dnaB and ATP, a signiicant amount of small fragment
was released by unwinding (lanes 6 and 10). For substrate 3, only the 3ʹ
fragment was unwound (lane 10). All unwinding was absolutely dependent on ATP hydrolysis.
Unwinding was considerably enhanced by adding single-strand
DNA-binding protein (SSB) (compare Figure 5–7, lanes 5 and 6 and lanes
9 and 10). Interestingly, SSB had to be added about 3 minutes after dnaB;
otherwise it inhibited unwinding.
A. Why do you suppose ATP hydrolysis is required for unwinding?
B. In what direction does dnaB move along the long single-strand DNA? Is
this direction more consistent with its movement on the template for the
leading strand or on the template for the lagging strand at a replication
fork?
C. Why do you suppose SSB inhibits unwinding when it is added before
dnaB, but stimulates unwinding when added after dnaB?
base-paired region
3′
5′
substrate 1
single-stranded tails
3′
5′
substrate 2
3′
5′
3′
5′
5′
3′
substrate 3
Figure 5–6 Substrates used to test the
properties of dnaB (Problem 5–36).
he diferent ways in which DNA synthesis occurs on the leading and
lagging strands raises the question as to whether synthesis occurs with
equal idelity on the two strands. One clever approach used reversion of
speciic mutations in the E. coli LacZ gene to address this question. E. coli
DnaB
SSB
substrate 1
substrate 2
substrate 3
+ +
+
+ +
+
+ +
+
+
heated to 100˚C
+
+
direction of
migration
labeled
fragments bound
to unlabeled DNA
free singlestranded labeled
DNA molecules
3′ fragment
5′ fragment
1
2
3
4
5
6
7
8
9 10 11 12
Figure 5–7 Results of several experiments
to measure unwinding by dnaB (Problem
5–36). only the single-strand fragments
were radioactively labeled. Their positions
are shown by the bands in this schematic
diagram. ATP was included with dnaB in all
incubations, which were carried out at 37°C.
84
Chapter 5: DNA Replication, Repair, and Recombination
(A)
5′
(B)
A
AG
TA C
lagging
GT
T
AT
TT G
LacZ
TC
lagging
R orientation
AA
AC
L
OriC
5′
L orientation
3′
R
3′
TT
AG
TG
CA
5′
leading
AT G
TC
T
AA C
TA
A
5′
leading
3′
3′
is a good choice for such a study because the same polymerase (DNA pol
III) synthesizes both the leading and the lagging strands.
he LacZ CC106 allele can regain its function (revert) by converting
the mutant A-T base pair to the normal G-C base pair. his allele was
inserted into the E. coli chromosome on one side of the normal origin
of replication (Figure 5–8A). Two E. coli strains were isolated: one with
the allele in the “L” orientation, and the other with it in the opposite, “R,”
orientation. As shown in Figure 5–8B, misincorporation of G opposite T
on one strand, or of C opposite A on the other strand, could lead to reversion. Previous studies had shown that C is very rarely misincorporated
opposite A. hus, the most common source of reversion is from misincorporation of G opposite T.
To eliminate the complicating efects of mismatch repair, the experiments were done in two mutant strains of bacteria. One was defective for
mismatch repair, which eliminates it from consideration; the other was
defective in the proofreading exonuclease, and introduces so many mismatches that it overwhelms the mismatch-repair machinery.
Accurate frequencies of LacZ reversion were measured in the two
strains, along with the frequencies of mutation at the Rif gene, whose orientation in the chromosome was constant (Table 5–2).
A. On which strand, leading or lagging, does DNA synthesis appear to be
more accurate? Explain your reasoning.
B. Can you suggest a reason why DNA synthesis might be more accurate on
the strand you have chosen?
TABLE 5–2 Frequencies (per 106 cells) of revertants of LacZ and mutants of Rif in
strains of E. coli deficient for mismatch repair or proofreading (Problem 5–37).
LacZ allele
(mutation
measured)
LacZ
orientation
CC106
(AT → GC)
L
0.27
7.0
2.7
82
CC106
(AT → GC)
R
0.51
6.4
6.7
80
Mismatch-repair deicient
Proofreading deicient
Lac– → Lac+ Rif s → Rif r
Lac– → Lac+ Rif s → Rif r
THE INITIATION AND COMPLETION OF DNA
REPLICATION IN CHROMOSOMES
TERMS TO LEARN
histone chaperone
origin recognition complex (ORC)
replication origin
S phase
telomerase
Figure 5–8 Fidelity of synthesis of the
leading and lagging strands (Problem
5–37). (A) Site of insertion of LacZ into
the E. coli chromosome. orientations
are denoted R and l. The arrows at
oriC represent the two forks initiated at
that site. (B) Arrangement of sequences
relative to the leading and lagging strands
in the l and R orientations. The G and
C misincorporations that could lead to
reversion are indicated.
THE INITIATION AND COMPLETION OF DNA REPLICATION IN CHROMOSOMES
85
DEFINITIONS
Match each deinition below with its term from the list above.
5–38
Period during a eukaryotic cell cycle in which DNA is synthesized.
5–39
Large multimeric protein structure that is bound to the DNA at origins of
replication in eukaryotic chromosomes throughout the cell cycle.
5–40
Special DNA sequence on a bacterial or viral chromosome at which DNA
replication begins.
5–41
Enzyme that elongates telomeres, the repetitive nucleotide sequences
found at the ends of eukaryotic chromosomes.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
5–42
In a replication bubble, the same parental DNA strand serves as the template strand for leading-strand synthesis in one replication fork and as
the template for lagging-strand synthesis in the other fork.
5–43
When bidirectional replication forks from adjacent origins meet, a leading strand always runs into a lagging strand.
5–44
If an origin of replication is deleted from a eukaryotic chromosome, the
DNA on either side will ultimately be lost, as well, because it cannot be
replicated.
THOUGHT PROBLEMS
5–45
5–46
he laboratory you joined is studying the life cycle of an animal virus that
uses circular, double-strand DNA as its genome. Your project is to deine
the location of the origin(s) of replication and to determine whether replication proceeds in one or both directions away from an origin (unidirectional or bidirectional replication). To accomplish your goal, you isolated replicating molecules, cleaved them with a restriction nuclease that
cuts the viral genome at only one site to produce a linear molecule from
the circle, and examined the resulting molecules in the electron microscope. Some of the molecules you observed are illustrated schematically
in Figure 5–9. (Note that it is impossible to distinguish the orientation of
one DNA molecule from another in the electron microscope.)
You must present your conclusions to the rest of the lab tomorrow.
How will you answer the two questions your advisor had posed for you?
Is there a single, unique origin of replication or several origins? Is replication unidirectional or bidirectional?
Which one of the following statements about the newly synthesized
strand of a human chromosome is correct?
A. It was synthesized from a single origin solely by continuous DNA synthesis.
B. It was synthesized from a single origin solely by discontinuous DNA synthesis.
C. It was synthesized from a single origin by a mixture of continuous and
discontinuous DNA synthesis.
D. It was synthesized from multiple origins solely by continuous DNA synthesis.
E. It was synthesized from multiple origins solely by discontinuous DNA
synthesis.
F. It was synthesized from multiple origins by a mixture of continuous and
discontinuous DNA synthesis.
original molecule
bubbles
“H”-forms
Figure 5–9 Parental and replicating forms
of an animal virus (Problem 5–45).
86
Chapter 5: DNA Replication, Repair, and Recombination
G. It was synthesized from multiple origins by either continuous or discontinuous DNA synthesis, depending on which speciic daughter chromosome is being examined.
5–47
he mechanism of DNA replication gives rise to the “end-replication
problem” for linear chromosomes. Over time, this problem leads to
loss of DNA from the ends of chromosomes. In cells such as yeast, loss
of nucleotides during replication is balanced by addition of nucleotides by telomerase. In humans, however, telomerase is turned of in
most somatic cells early in development, so that chromosomes become
shorter with increasing rounds of replication. Consider one round of replication in a human somatic cell. Which one of the following statements
correctly describes the status of the two daughter chromosomes relative
to the parent chromosome?
A. One daughter chromosome will be shorter at one end; the other daughter chromosome will be normal at both ends.
B. One daughter chromosome will be shorter at both ends; the other daughter chromosome will be normal at both ends.
C. One daughter chromosome will be shorter at both ends; the other daughter chromosome will be shorter at only one end.
D. Both daughter chromosomes will be shorter at one end, which is the
same end in the two chromosomes.
E. Both daughter chromosomes will be shorter at one end, which is the
opposite end in the two chromosomes.
F. Both daughter chromosomes will be shorter at both ends.
CALCULATIONS
5–48
In the early embryo of Drosophila, many replication origins are active so
that several can be observed in a single electron micrograph, as shown in
Figure 5–10.
0.1 µm
A. Identify the four replication bubbles in Figure 5–10. Indicate the approximate locations of the origins at which each replication bubble was initiated, and label the replication forks 1 through 8 from left to right across
the igure.
B. Estimate how long it will take until forks 4 and 5 collide with each other.
How long will it take until forks 7 and 8 collide? Each nucleotide in DNA
occupies 0.34 nm, and eukaryotic replication forks move at about 50
nucleotides/second. For this problem disregard the nucleosomes evident in Figure 5–10 and assume that the DNA is fully extended.
5–49
Assuming that there were no time constraints on replication of the
genome of a human cell, what would be the minimum number of origins that would be required? If replication had to be accomplished in an
8-hour S phase and replication forks moved at 50 nucleotides/second,
what would be the minimum number of origins required to replicate the
human genome? (Recall that the human genome comprises a total of
6.4 × 109 nucleotides on 46 chromosomes.)
DATA HANDLING
5–50
You are investigating DNA synthesis in tissue-culture cells, using 3H-thymidine to radioactively label the replication forks. By breaking open the
Figure 5–10 Electron micrograph showing
four replication bubbles in a chromosome
from the early embryo of Drosophila
(Problem 5–48).
THE INITIATION AND COMPLETION OF DNA REPLICATION IN CHROMOSOMES
Figure 5–11 Autoradiographic
investigation of DNA replication in
cultured cells (Problem 5–50). (A) Addition
of 3H-labeled thymidine immediately after
release from the synchronizing block.
(B) Addition of 3H-labeled thymidine
30 minutes after release from the
synchronizing block.
(A)
(B)
50 µm
cells in a way that allows some of the DNA strands to be stretched out,
very long DNA strands can be isolated intact and examined. You overlay
the DNA with a photographic emulsion, and expose it for 3 to 6 months, a
procedure known as autoradiography. Because the emulsion is sensitive
to radioactive emissions, the 3H-labeled DNA shows up as tracks of silver
grains. Because the stretching collapses replication bubbles, the daughter duplexes lie side by side and cannot be distinguished from each other.
You pretreat the cells to synchronize them at the beginning of
S phase. In the irst experiment, you release the synchronizing block and
add 3H-thymidine immediately. After 30 minutes, you wash the cells and
change the medium so that the total concentration of thymidine is the
same as it was, but only one-third of it is radioactive. After an additional
15 minutes, you prepare DNA for autoradiography. he results of this
experiment are shown in Figure 5–11A. In the second experiment, you
release the synchronizing block and then wait 30 minutes before adding 3H-thymidine. After 30 minutes in the presence of 3H-thymidine, you
once again change the medium to reduce the concentration of radioactive thymidine and incubate the cells for an additional 15 minutes. he
results of the second experiment are shown in Figure 5–11B.
A. Explain why, in both experiments, some regions of the tracks are dense
with silver grains (dark), whereas others are less dense (light).
B. In the irst experiment, each track has a central dark section with light
sections at each end. In the second experiment, the dark section of each
track has a light section at only one end. Explain the reason for this diference.
C. Estimate the rate of fork movement (μm/min) in these experiments. Do
the estimates from the two experiments agree? Can you use this information to gauge how long it would take to replicate the entire genome?
5–51
87
Segments of DNA that allow plasmids to replicate in yeast are known as
autonomous replication sequences (ARSs). hey are thought to function
as origins of replication. Proving that an ARS is an origin of replication is
diicult, however, mainly because it is very hard to obtain enough welldeined replicating DNA to analyze. his problem can be addressed using
a two-dimensional gel-electrophoretic analysis that separates DNA molecules by mass in the irst dimension and by shape in the second dimension. Because they have branches, replicating molecules migrate more
slowly in the second dimension than do linear molecules of equal mass.
By cutting replicating molecules with restriction nucleases, it is possible
to generate a continuum of diferent branched forms that together give
characteristic patterns on two-dimensional gels (Figure 5–12).
You apply this technique to the replication of a plasmid that contains
a speciic ARS, Ars1. To maximize the fraction of plasmid molecules that
are replicating, you synchronize a yeast culture and isolate DNA from
cells in S phase. You then digest the DNA with BglII or PvuI, which cut the
plasmid as indicated in Figure 5–13A. You separate the DNA fragments
88
Chapter 5: DNA Replication, Repair, and Recombination
(A) ONE
BRANCH
(B) SYMMETRIC
BUBBLE
(C) TWO
BRANCHES
(D) ASYMMETRIC
BUBBLE
second dimension
(shape)
first dimension (mass)
2 kb
2 kb
1 kb
2 kb
1 kb
Figure 5–12 Expected patterns on
two-dimensional gels for various DNA
molecules (Problem 5–51). (A) Molecules
with a single branch. (B) Molecules with a
symmetrically located replication bubble.
(C) Molecules with two branches.
(D) Molecules with an asymmetrically
located replication bubble. In the upper
half of the igure, 1-kb molecules
are shown at progressive stages of
replication to 2-kb molecules. In the
lower half are shown the corresponding
gel patterns that would result from a
continuum of such intermediates.
2 kb
1 kb
1 kb
by two-dimensional electrophoresis and visualize the plasmid sequences
by autoradiography after blot hybridization to radioactive plasmid DNA
(Figure 5–13B).
A. What is the source of the intense spot of hybridization at the 4.5-kb position in both gels in Figure 5–13B?
B. Do the results of this experiment indicate that Ars1 is an origin of replication? Explain your answer.
C. here is a gap in the arc of hybridization in the PvuI gel pattern in Figure
5–13B. What is the basis for this discontinuity?
(B) GEL PATTERNS
first
second
(A) PLASMID
BglII
PvuI
Figure 5–13 Analysis of replication of an
Ars1-containing plasmid (Problem 5–51).
(A) Structure of the Ars1 plasmid. (B) The
two-dimensional gel patterns resulting
from cleavage of replicating plasmids
with BglII or PvuI.
PvuI
9 kb
Ars1
BglII
5–52
4.5 kb
he shell of a Drosophila egg is made from more than 15 diferent chorion proteins, which are synthesized at a late stage in egg development by
follicle cells surrounding the egg. he various chorion genes are grouped
in two clusters, one on chromosome 3 and the other on the X chromosome. In each cluster, the genes are closely spaced with only a few
hundred nucleotides separating adjacent genes. During egg development, the number of copies of the chorion genes increases by overreplication of a segment of the surrounding chromosome. he level of ampliication around the chorion cluster on chromosome 3 is maximal in the
region of the chorion genes, but extends for nearly 50 kb on each side
(Figure 5–14).
he DNA sequence responsible for ampliication of the cluster on
chromosome 3 has been narrowed to a 510-nucleotide segment immediately upstream of one of the chorion genes. When this segment is moved
to diferent places in the genome, those new sites are also ampliied in
chorion gene cluster
fold amplification
4.5 kb
60
40
20
0
0
20
40
60
80
100
kilobases
Figure 5–14 levels of ampliication in the
region of the chromosome surrounding
the chorion gene cluster (Problem 5–52).
THE INITIATION AND COMPLETION OF DNA REPLICATION IN CHROMOSOMES
5–53
5–54
5–55
In yeast, origin selection is initiated by the origin recognition complex
(ORC). ORC is a six-protein DNA-binding complex that recognizes DNA
sequences within the yeast origin of replication. One of the components
of ORC, Orc1, contains a protein motif (the Walker motif ) that is commonly associated with binding and hydrolysis of ATP.
To determine whether binding of ATP, its hydrolysis, or both are
required for the recognition of origin DNA by ORC, you carry out the following set of experiments. You irst mutate the Orc1 gene to change an
amino acid in the Walker motif of Orc1. You then isolate two versions
of ORC: the wild-type form with normal Orc1 and the mutant form with
a defective Walker motif. Finally, you measure the binding of these two
forms of ORC to origin DNA in the presence of diferent concentrations of
ATP. Binding of the two forms of ORC, as revealed by a DNase I protection
assay (DNase footprinting), is shown in Figure 5–15. Exactly the same
footprints were obtained when the nonhydrolyzable analog, ATPγS, was
used in place of ATP.
A. Indicate the location of ORC binding on the origin DNA in Figure 5–15.
B. Is ATP required for ORC to bind to origin DNA? How can you tell?
C. Is ATP hydrolysis required for ORC binding to origin DNA? How can you
tell?
D. Is the Walker motif important to the function of ORC? Explain your
answer.
You have developed an assay for assembly of nucleosomes onto DNA in
order to deine the role of chromosome assembly factor 1 (CAF1). You
replicate the circular genome of the SV40 animal virus in a cell-free system in the presence of a labeled nucleotide to tag the replicated molecules, so you can follow them speciically in the assembly assay. After
separating the DNA from soluble components in the cell-free replication
system, you incubate it with puriied CAF1 and a source of histones. You
then assay for nucleosome assembly by its efects on the supercoiling status of the circular SV40 genome. Genomes without nucleosomes remain
relaxed, whereas genomes with nucleosomes become supercoiled. You
separate diferent supercoiled forms of the DNA by electrophoresis on an
agarose gel, which was stained to reveal all forms of DNA and subjected
to autoradiography to identify replicated DNA (Figure 5–16).
A. Is most of the SV40 DNA replicated or unreplicated? How can you tell?
B. Does CAF1 assemble nucleosomes on replicated DNA, unreplicated
DNA, or both? Explain your answer.
C. How do you suppose that CAF1 recognizes the DNA it assembles into
nucleosomes?
You have recently puriied and partially sequenced a protein from a ciliated protozoan that seems to be the catalytic subunit of telomerase. You
then identify the homologous gene in ission yeast, so you can perform
genetic studies that are impossible in the protozoan. You make a targeted
deletion of one copy of the gene in a diploid strain of the yeast and then
induce sporulation to produce haploid organisms. All four spores germinate perfectly, and you are able to grow colonies on nutrient agar plates.
Every 3 days, you re-streak colonies onto fresh plates. After four such
serial transfers, the descendants of two of the original four spores grow
ORC
wild type
mutant
ATP 0
1
10 mM
follicle cells. No RNA or protein product seems to be synthesized from
this ampliication-control element.
A. Sketch what you think the DNA from an ampliied cluster would look like
in the electron microscope.
B. How many rounds of replication would be required to achieve a 60-fold
ampliication?
C. How do you think the 510-nucleotide ampliication-control element promotes the overreplication of the chorion gene cluster?
89
00
2
3
4
5
6
7
8
9
0
11 13 15
10 12 14
Figure 5–15 DNase footprinting assay
to detect oRC binding to origin DNA
(Problem 5–53). Wedges indicate increasing
concentration of ATP in factors of ten, from
10 nM to 100 μM. In the footprinting assay,
one strand of the origin-containing DNA
was labeled at one end. After oRC was
allowed to bind, the complex was treated
briely with DNase I, which breaks the DNA
at characteristic places, except where it is
protected by oRC. lanes 7 and 15 omit
both ATP and oRC.
CAF1
+
CAF1
+
relaxed
supercoiled
DNA stain
autoradiogram
Figure 5–16 CAF1 nucleosome-assembly
assay (Problem 5–54). The presence or
absence of CAF1 in the assembly assay
is indicated by + or –, respectively.
90
Chapter 5: DNA Replication, Repair, and Recombination
days
spore 1 spore 2 spore 3 spore 4
3 6 9 3 6 9 3 6 9 3 6 9
markers (bp)
600
500
400
300
200
100
poorly, if at all. You take cells from the 3-, 6-, and 9-day master plates,
prepare DNA from them, and cleave the samples at a chromosomal site
about 35 nucleotides away from the start of the telomere repeats. You
separate the fragments by gel electrophoresis, and hybridize them to a
radioactive telomere-speciic probe (Figure 5–17).
A. What is the average length of telomeres in ission yeast?
B. Do the data support the idea that you have identiied yeast telomerase? If
so, which spores lack telomerase?
C. Assuming that the generation time of this yeast is about 6 hours when
growing on plates, by how much do the chromosomes shorten in each
generation in the absence of telomerase?
D. If you were to examine the yeast cells that stop dividing, do you suppose
they would be smaller, larger, or about the same size as normal yeast
cells?
DNA REPAIR
TERMS TO LEARN
base excision repair
DNA repair
nonhomologous end joining
nucleotide excision repair
DEFINITIONS
Match each deinition below with its term from the list above.
5–56
A means for repairing double-strand DNA breaks that links two ends with
little regard for sequence homology.
5–57
Collective term for the enzymatic processes that correct deleterious
changes afecting the continuity or sequence of a DNA molecule.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
5–58
DNA repair mechanisms all depend on the existence of two copies of the
genetic information, one in each of the two homologous chromosomes.
5–59
Spontaneous depurination and the removal of a deaminated C by uracil
DNA glycosylase leave identical substrates, which are recognized by AP
endonuclease.
5–60
Only the initial steps in DNA repair are catalyzed by enzymes that are
unique to the repair process; the later steps are typically catalyzed by
enzymes that play more general roles in DNA metabolism.
THOUGHT PROBLEMS
5–61
Discuss the following statement: “he DNA repair enzymes that correct
Figure 5–17 Analysis of telomeres from
four ission-yeast spores (Problem 5–55).
The results from normal diploid yeast
are shown on the right, adjacent to the
markers.
DNA REPAIR
91
damage introduced by deamination and depurination must preferentially recognize such defects on newly synthesized DNA strands.”
5–62
If you compare the frequency of the sixteen possible dinucleotide
sequences in the E. coli and human genomes, there are no striking diferences except for one dinucleotide, 5ʹ-CG-3ʹ. he frequency of CG dinucleotides in the human genome is signiicantly lower than in E. coli and
signiicantly lower than expected by chance. Why do you suppose that
CG dinucleotides are underrepresented in the human genome?
CALCULATIONS
5–63
Ku70 and Ku80, two key proteins used in nonhomologous end joining
(NHEJ), form heterodimers that bind to broken DNA ends, helping to
align them for joining. How diicult is it for a Ku dimer to ind a doublestrand break? One way to approach this question is to estimate the average distance between Ku dimers in the nucleus: a break in the DNA will
be within half that distance from a Ku dimer. If there are 4 × 105 Ku dimers in a typical nucleus, what is their average separation? Assume that
the Ku dimers are randomly distributed, that a Ku dimer can be approximated as a cube, 8 nm on a side, and that the nucleus is 6 μm in diameter.
[Volume of a sphere is (4/3)πr3; volume of a cube is l 3.]
5–64
With age, somatic cells are thought to accumulate genomic “scars” as a
result of the inaccurate repair of double-strand breaks by nonhomologous end joining (NHEJ). Estimates based on the frequency of breaks in
primary human ibroblasts suggest that by age 70, each human somatic
cell may carry some 2000 NHEJ-induced mutations due to inaccurate
repair. If these mutations were distributed randomly around the genome,
how many protein-coding genes would you expect to be afected? Would
you expect cell function to be compromised? Why or why not? (Assume
that 2% of the genome—1.5% protein-coding and 0.5% regulatory—is
crucial information.)
DATA HANDLING
Several genes in E. coli, including UvrA, UvrB, UvrC, and RecA, are
involved in repair of UV damage. Strains of E. coli that are defective in
any one of these genes are much more sensitive to killing by UV light than
are nonmutant (wild-type) cells, as shown for UvrA and RecA strains
in Figure 5–18A. Cells mutant for pairs of these genes display a wide
range of sensitivity to UV light. he combinations of Uvr mutations with
(A)
(B)
wild type
100
100
37
37
10
10
UvrARecA
cell survival (%)
5–65
UvrA
1.0
1.0
RecA
0.1
0.1
UvrARecA
0.01
0.01
0
5
UV dose (J/m2)
10
0.1
0.2
UV dose (J/m2)
Figure 5–18 Cell survival as a function of
UV dose (Problem 5–65). (A) Survival of
wild-type cells, a UvrA mutant, a RecA
mutant, and a UvrARecA double mutant.
(B) An expanded scale for UvrARecA
survival.
Chapter 5: DNA Replication, Repair, and Recombination
% of total radioactivity
one another show little increase in sensitivity relative to the Uvr single
mutants. By contrast, the combination of a RecA mutation with any of the
Uvr mutations gives a strain that is exquisitely sensitive to UV light, as
shown for UvrARecA on an expanded scale in Figure 5–18B.
A. Why do combinations of a RecA mutation with a Uvr mutation give an
extremely UV-sensitive strain of bacteria, whereas combinations of mutations in diferent Uvr genes are no more UV-sensitive than the individual
mutants?
B. According to the Poisson distribution, when a population of bacteria
receives an overall average of one lethal “hit,” 37% (which is e–1) will survive because they receive no hits. For the double mutant UvrARecA, a UV
dose of 0.04 joules/m2 gives 37% survival (Figure 5–18B). Calculate the
number of pyrimidine dimers that constitutes a lethal hit for the UvrARecA strain. E. coli has 4.6 × 106 base pairs in its genome (assume 50% GC).
Exposure of DNA to UV light at 400 joules/m2 converts 1% of the total
pyrimidine pairs (TT, TC, CT, plus CC) to pyrimidine dimers.
5–66
he nature of the mutagenic lesion introduced by the alkylating agent
MNNG (N-methyl-Nʹ-nitro-N-nitrosoguanidine) and the mechanism
of its removal from DNA were identiied in the following experiments.
To determine the nature of the mutagenic lesion, untreated bacteria and
bacteria that had been exposed to low doses of MNNG were incubated
with 50 μg/mL 3H-MNNG for 10 minutes. Bacteria exposed briely to a
low dose of MNNG induce a repair enzyme that removes methyl groups
from DNA, allowing these bacteria to survive better and sufer fewer
mutations than bacteria that were not pre-treated with MNNG. DNA was
isolated from bacteria that were not pre-treated with MNNG and from
those that were adapted by exposure to a low concentration of MNNG.
he DNAs were hydrolyzed to nucleotides, and the radioactive purines
were then analyzed by paper chromatography as shown in Figure 5–19.
he mechanism of removal of the mutagenic lesion was investigated
by irst purifying the enzyme responsible. he kinetics of removal were
studied by incubating diferent amounts of the enzyme (molecular
weight 19,000) with DNA containing 0.26 pmol of the mutagenic base,
which was radioactively labeled with 3H. At various times, samples were
taken, and the DNA was analyzed to determine how much of the mutagenic base remained (Figure 5–20). When the experiment was repeated
at 5°C instead of 37°C, the initial rates of removal were slower, but exactly
the same end points were achieved.
A. Which methylated purine is responsible for the mutagenic action of
MNNG?
B. What is peculiar about the kinetics of removal of the methyl group from
the mutagenic base? Is this peculiarity due to an unstable enzyme?
C. Calculate the number of methyl groups that are removed by each enzyme
molecule. Does this calculation help to explain the peculiar kinetics?
40
30
7-methylguanine
3-methyladenine
O6-methylguanine
20
100
% radioactivity remaining
92
1.25 ng
75
2.50 ng
50
25
10
5.00 ng
0
origin
5
distance (cm)
10
Figure 5–19 Chromatographic separation of labeled methylated purines in the DNA of
untreated bacteria and bacteria treated with low doses of MNNG (Problem 5–66). The
red line indicates methylated purines from bacteria that were not pre-treated with a
low dose of MNNG; the blue line indicates methylated purines from bacteria that were
pre-treated with a low dose of MNNG.
0
0
2
4
time (minutes)
6
Figure 5–20 Removal of 3H-labeled
methyl groups from DNA by the
puriied enzyme (Problem 5–66). The
quantities of puriied enzyme are
indicated.
DNA REPAIR
5–67
93
0
2
—GCAATC GTGGAG—
—CGTTAG CACCTC—
sequence A
—ATCCG GT TTCGA—
—TAGGC CA AAGCT—
—GCAATC AATCCG—
—CGTTAG TTAGGC—
rearranged
sequence
—ATCCG GT ACCAG—
—TAGGC CA TGGTC—
—CCGTAG AATCCG—
—GGCATC TTAGGC—
sequence B
—CCTTA GT ACCAG—
—GGAAT CA TGGTC—
Figure 5–21 Microhomology at
rearrangement junctions (Problem 5–67).
In the rearrangement junction with two
nucleotides of microhomology, the central
GT cannot be uniquely assigned to either
sequence A or sequence B; it could have
come from either.
Nonhomologous end joining (NHEJ) is responsible for linking DNA
sequences that are not homologous to one another. Junctions formed
by NHEJ are commonly found at sites of DNA rearrangements, including translocations, inversions, duplications, and deletions. Surveys of
large numbers of such junctions have revealed the presence of so-called
“microhomology” at the junctions, typically ranging from 0–5 nucleotides. Examples of junctions with 0 and 2 nucleotides of microhomology are shown in Figure 5–21.
Are these short homologies relevant to the mechanism of NHEJ, or
are they present by chance? One way to decide the issue is to compare
the distribution of observed junctional microhomology with the distribution expected by chance. Microhomologies at 110 NHEJ junctions
are shown along with the distribution expected by chance in Table 5–3.
One way to determine whether these two distributions are diferent is to
use the statistical method known as chi-square analysis. he readout of
a chi-square analysis is a P value, which measures the probability that
the observed distribution is the same as the expected distribution. Commonly, a P value of less than 0.05 is taken to mean that the two distributions are diferent. (A P value of 0.05 means that there is a 5% chance that
the observed distribution is the same as the expected distribution.)
Using chi-square analysis, decide whether the observed distribution
in Table 5–3 is the same as or diferent from the distribution expected by
chance. If you don’t have a statistics package available, search for “chi
square calculator” on the Internet.
TABLE 5–3 Observed distribution of microhomologies at
rearrangement junctions along with the distribution expected by
chance (Problem 5–67).
Microhomology
Distribution
0
1
2
3
4
5
Observed
47
21
14
13
9
6
Expected
62
31
12
4
1
0
The distribution of microhomology expected by chance was calculated
on the assumption that the random joining of blunt-ended DNA segments
generated the junctions.
MEDICAL LINKS
5–68
Humans with the rare genetic disease xeroderma pigmentosum (XP)
are extremely sensitive to sunlight and are prone to developing malignant skin cancers. Defects in any one of eight genes can cause XP.
Seven XP genes encode proteins involved in nucleotide excision repair
(NER). he eighth gene is associated with the XP variant (XP-V) form of
the disease. Cells from XP-V patients are proicient for NER but do not
94
Chapter 5: DNA Replication, Repair, and Recombination
(A) SUBSTRATE
5ʹ [32P] – CACTGACTGTATG
GTGACTGACATACTACTTCTACGACTGCTC – 5ʹ
(B) RESULTS
XP-V enzyme
polymerase α
no
template
– – +
– + –
no
damage
cyclobutane 6-4 photodimer
product
– – + + – – + + – – + +
– + – + – + – + – + – +
30
17
16
13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
accurately replicate UV-damaged DNA. Using a clever assay, you manage
to purify from a normal cell extract the enzyme that is missing from XP-V
cells. You test its ability to synthesize DNA from the simple template
in Figure 5–22A, which contains a TT sequence. his template can be
modiied to contain either a cyclobutane thymine dimer or the somewhat rarer 6-4 photoproduct (two Ts linked in a diferent way). You compare the ability of DNA polymerase α (a normal replicative polymerase)
and the XP-V enzyme to synthesize DNA from the undamaged template,
the template with a cyclobutane dimer, and the template with a 6-4 photoproduct. By labeling the primer (the shorter DNA strand) at its 5ʹ end
and denaturing the reaction products, it is possible to determine whether
DNA synthesis has occurred (Figure 5–22B).
A. Is the XP-V enzyme a DNA polymerase? Why or why not?
B. How does the XP-V enzyme difer from DNA polymerase α on an undamaged template? On a template with a cyclobutane dimer? On a template
with a 6-4 photoproduct?
C. How accurately do you suppose that the XP-V enzyme copies normal
DNA? Would you guess it to be error-prone or faithful?
D. If NER is normal in patients with XP-V, why are they sensitive to sunlight
and prone to skin cancers?
5–69
Mutagens such as N-methyl-Nʹ-nitro-N-nitrosoguanidine (MNNG) and
methyl nitrosourea (MNU) are potent DNA-methylating agents and are
extremely toxic to cells. Nitrosoguanidines are used in research as mutagens and clinically as drugs in cancer chemotherapy because they preferentially kill cells in the act of replication.
he original experiment that led to the discovery of the alkylation
repair system in bacteria was designed to assess the long-term efects
of exposure to low doses of MNNG (as in chemotherapy), in contrast to
brief exposure to large doses (as in mutagenesis). Bacteria were placed
irst in a low concentration of MNNG (1 μg/mL) for 1.5 hours and then
in fresh medium lacking MNNG. At various times during and after exposure to the low dose of MNNG, samples of the culture were treated with a
high concentration of MNNG (100 μg/mL) for 5 minutes and then tested
for viability and the frequency of mutants. As shown in Figure 5–23,
Figure 5–22 Comparison of DNA
polymerase α and the XP-V enzyme
(Problem 5–68). (A) Substrate for DNA
polymerization. The adjacent Ts indicated
by the bracket can be selectively modiied
to produce a cyclobutane dimer or a 6-4
photoproduct. (B) Results obtained with
DNA polymerase α and the XP-V enzyme
on damaged and undamaged templates.
Numbers on the left indicate lengths of
single-strand DNA in nucleotides.
HOMOLOGOUS RECOMBINATION
50
% survival
100
50
0
25
–2
–1
0
1
2
3
4
0
time (hours)
Figure 5–23 Adaptive response of E. coli to low doses of MNNG (Problem 5–69).
MNNG at 1 μg/ml was present from –1.5 to 0 hours, as indicated by arrows.
Samples were removed at various times and treated briely with 100 μg/ml MNNG to
assess the number of survivors and the frequency of mutants.
exposure to low doses of MNNG temporarily increased the number of
survivors and decreased the frequency of mutants among the survivors.
As shown in Figure 5–24, the adaptive response to low doses of MNNG
was prevented if chloramphenicol (an inhibitor of protein synthesis) was
included in the incubation.
A. Does the adaptive response of E. coli to low levels of MNNG require activation of preexisting protein or the synthesis of new protein?
B. Why do you think the adaptive response to a low dose of MNNG is so
short-lived?
HOMOLOGOUS RECOMBINATION
TERMS TO LEARN
allele
gene conversion
Holliday junction
homologous recombination
hybridization
loss of heterozygosity
Rad51
RecA
strand exchange
DEFINITIONS
Match each deinition below with its term from the list above.
5–70
One of a set of alternative forms of a gene. In a diploid cell, each gene will
have two of these, each occupying the same position (locus) on homologous chromosomes.
5–71
Process by which DNA sequence information can be transferred from
one DNA helix (which remains unchanged) to another DNA helix whose
sequence is altered.
5–72
Experimental process in which two complementary nucleic acid strands
form a double helix; a powerful technique for detecting speciic nucleotide sequences.
5–73
X-shaped structure observed in DNA undergoing recombination, in
which the two DNA molecules are held together at the site of crossingover, also called a cross-strand exchange.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
5–74
Homologous recombination requires relatively long regions of homologous DNA on both partners in the exchange.
mutants per million survivors
remove
MNNG
mutants per million survivors
add
MNNG
95
50
+ chloramphenicol
40
30
20
10
0
– chloramphenicol
0
20
40
60
80
exposure to MNNG (minutes)
Figure 5–24 Effects of chloramphenicol
on the adaptive response to low doses
of MNNG (Problem 5–69). After different
times of exposure to 1 μg/ml MNNG,
samples were removed and treated
with 100 μg/ml MNNG to measure
susceptibility to mutagenesis.
96
Chapter 5: DNA Replication, Repair, and Recombination
5–75
Gene conversion requires a limited amount of DNA synthesis.
(A)
A
B
C
D
A
D
C
B
A
B
C
D
A
B
C
D
THOUGHT PROBLEMS
5–76
5–77
5–78
5–79
Using Figure 5–25A as a guide, draw the products of a crossover recombination event between the homologous regions of the molecules represented in Figure 5–25B. In the igure, single lines represent the DNA
duplex and arrows represent the targets for homologous recombination.
Haploid yeast cells that preferentially repair double-strand breaks by
homologous recombination are especially sensitive to agents that cause
double-strand breaks in DNA. If the breaks occur in the G1 phase of the
cell cycle, most yeast cells die; however, if the breaks occur in the G2
phase, a much higher fraction of cells survive. Explain these results.
When plasmid DNA is extracted from E. coli and examined by electron
microscopy, the majority of molecules are monomeric circles, but there
are a variety of other forms, including dimeric and trimeric circles. In
addition, about 1% of the molecules appear as igure-8 forms, in which
the two loops are equal (Figure 5–26A).
You suspect that the igure-8 forms are recombination intermediates in the formation of a dimer from two monomers (or two monomers from a dimer). To rule out the possibility that they might represent twisted dimers or touching monomers, you digest the DNA sample
with a restriction nuclease that cuts at a single site in the monomer, and
then examine the molecules. After cutting, only two forms are seen: 99%
of the DNA molecules are linear monomers, and 1% are χ (chi) forms
(Figure 5–26B). You note that the χ forms have an interesting property:
the two longer arms are the same length, as are the two shorter arms. In
addition, the sum of the lengths of a long arm and a short arm is equal to
the length of the monomer plasmid. he position of the crossover point,
however, is completely random.
Unsure of yourself and feeling you are probably missing some
hidden artifact, you show your pictures to a friend. She points out that
your observations prove you are looking at recombination intermediates
that arose by random pairing at homologous sites.
A. Is your friend correct? What is her reasoning?
B. How would you expect your observations to difer if you repeated the
experiments in a strain of E. coli that carried a nonfunctional RecA gene?
C. How would the χ forms have difered from those you observed if the
igure-8 forms had arisen as intermediates in a site-speciic recombination between the monomers?
D. What would the χ forms have looked like if the igure-8 forms had arisen
as intermediates in a totally random, nonhomologous recombination
between the monomers?
(B)
1.
2.
A
B
C
D
A
B
D
C
3.
4.
A
B
C
D
A
B
D
C
5.
6.
Figure 5–25 A variety of recombination
substrates (Problem 5–76).
Discuss the following statement: “he Holliday junction contains two
distinct pairs of strands (crossing strands and noncrossing strands),
which cannot be interconverted without breaking the phosphodiester
backbone of at least one strand.”
(A) BEFORE CUTTING
(B) AFTER CUTTING
monomer
χ forms
figure-8 form (dimer)
dimer
linears
Figure 5–26 Structures of plasmid
molecules (Problem 5–78). (A) Before
digestion. (B) After digestion with a
single-cut restriction nuclease.
HOMOLOGOUS RECOMBINATION
5–80
5–81
97
Draw the structure of the double Holliday junction that would result from
strand invasion by both ends of the broken duplex into the intact homologous duplex shown in Figure 5–27. Label the left end of each strand in
the Holliday junction 5ʹ or 3ʹ so that the relationship to the parental and
recombinant duplexes is clear. Indicate how DNA synthesis would be
used to ill in any single-strand gaps in your double Holliday junction.
In addition to correcting DNA mismatches, the mismatch repair system functions to prevent homologous recombination from taking place
between similar but not identical sequences. Why would recombination
between similar, but nonidentical sequences pose a problem for human
cells?
DATA HANDLING
5–82
In E. coli, RecA protein catalyzes both the initial pairing step of recombination and subsequent branch migration. It promotes recombination by
binding to single-strand DNA and catalyzing the pairing of such coated
single strands to homologous double-strand DNA. One assay for the
action of RecA is the formation of double-strand DNA circles from a mixture of double-strand linear molecules and homologous single-strand
circles, as illustrated in Figure 5–28. his reaction proceeds in two steps:
circles pair with linears at an end and then they branch-migrate until a
single-strand linear DNA is displaced.
One important question about the RecA reaction is whether branch
migration is directional. his question has been studied in the following
way. Single-strand circles, which were uniformly labeled with 32P, were
mixed with unlabeled double-strand linear molecules in the presence of
RecA. As the single-strand DNA pairs with the linear DNA, it becomes
sensitive to cutting by restriction nucleases, which do not cut singlestrand DNA. By sampling the reaction at various times, digesting the
DNA with a restriction nuclease, and separating the labeled fragments by
electrophoresis, you obtain the pattern shown in Figure 5–29A.
single-stranded circle
+
–
double-stranded linear DNA
+
+
–
+
nicked double-stranded
circle
–
+
+
single-stranded linear DNA
Figure 5–28 Strand assimilation assay for
RecA (Problem 5–82). The single-strand
(+) circle is complementary to the minus
(–) strand and identical to the plus strand
of the duplex.
5′
3′
5′
3′
Figure 5–27 A broken duplex with singlestrand tails ready to invade an intact
homologous duplex (Problem 5–80).
Chapter 5: DNA Replication, Repair, and Recombination
98
(B) RESTRICTION MAP OF VIRAL DNA
(A) ELECTROPHORESIS OF LABELED
RESTRICTION FRAGMENTS
5′
3′
c
fragments
a
b
d
a
+ strand
c
Figure 5–29 Analysis of branch migration
catalyzed by RecA (Problem 5–82).
(A) Electrophoretic separation of labeled
restriction fragments as a function of
time of incubation with RecA. (B) Sites
of cleavage represented on the singlestrand DNA circle. Clockwise around the
circle is 5ʹ to 3ʹ.
d
b
e
e
f
f
0
2.5
5
7.5
10 12.5 15
20
25
30
time (minutes)
A. By comparing the time of appearance of labeled fragments with the
restriction map of the circular DNA in Figure 5–29B, deduce which end
(5ʹ or 3ʹ) of the minus strand of the linear DNA the circular plus strand
pairs with initially. Also deduce the direction of branch migration along
the minus strand. (he linear double-strand DNA was cut at the boundary between fragments a and c on the restriction map.)
B. Estimate the rate of branch migration, given that the length of this DNA is
7 kb.
C. What would you expect to happen if the linear double-strand DNA carried an insertion of 500 nonhomologous nucleotides between restriction
fragments e and a?
5–83
Most eukaryotic cells use two diferent mechanisms to repair doublestrand breaks in DNA: homologous recombination and nonhomologous
end joining. he recombination pathway is favored by a homolog of the
Rad52 protein, irst deined in yeast. Relatives of the protein known as Ku
stimulate the end-joining pathway.
You have puriied the human Rad52 protein and are studying its properties. When you mix Rad52 with linear DNA that has short (300-nucleotide) single-strand tails and examine the mixture in the electron microscope, you see structures like the one shown in Figure 5–30A. (Such
structures are much rarer when the linear DNA is blunt-ended.) You
then expose Rad52-bound, uniformly radiolabeled DNA to nucleases
and measure digestion of DNA by release of soluble fragments. As shown
in Figure 5–30B, Rad52 protects the DNA from exonuclease digestion but
not from digestion by an endonuclease.
A. Where does Rad52 bind on the linear DNA? What features of the DNA are
important for Rad52 binding?
B. How does Rad52 protect against digestion by the exonuclease, but not
against digestion by the endonuclease?
C. How do the properties of Rad52 revealed by these observations it into its
role in recombinational repair of double-strand breaks?
(B) NUCLEASE DIGESTION
exonuclease
(A) ELECTRON MICROGRAPH
acid-soluble 32P (%)
endonuclease
100
50
40
– Rad52
+/– Rad52
75
30
50
20
10
0
+ Rad52
0
10
20
30
time (minutes)
25
0
0
10
20
30
time (minutes)
Figure 5–30 Analysis of the role of Rad52
in repair of double-strand breaks (Problem
5–83). (A) Binding of Rad52 to linear DNA.
A schematic representation is shown
adjacent to the micrograph. (B) Nuclease
digestion of DNA in the presence (+) and
absence (–) of bound Rad52.
TRANSPOSITION AND CONSERVATIVE SITE-SPECIFIC RECOMBINATION
(A) SUBSTRATES
(B) RuvC ASSAYS
consensus
a
5′---G-A-T-T-G-C-T-A-G-G-C--- 3′
b
3′---C-T-A-A-C-G A-T-C-C-G--- 5′
c
3′---C-T-A-A-C-G A-T-C-C-G---5′ d
5′---G-A-T-T-G-C-T-A-G-G-C---3′
consensus mutant
hybrid
hybrid
RuvC
uncut
mutant
5′---G-G-T-T-G-C-T-A-G-G-C--- 3′
3′---C-C-A-A-C-G A-T-C-C-G--- 5′
cut
3′---C-C-A-A-C-G A-T-C-C-G--- 5′
5′---G-G-T-T-G-C-T-A-G-G-C--- 3′
hybrid
5′---G-A-T-T-G-C-T-A-G-G-C--- 3′
3′---C-T-A-A-C-G A-T-C-C-G--- 5′
3′---C-C-A-A-C-G A-T-C-C-G--- 5′
---G-G-T-T-G-C-T-A-G-G-C--- 3′
5′
5–84
he mechanism of Holliday junction cleavage by RuvC has been investigated using artiicial junctions created by annealing oligonucleotides
together (Figure 5–31A). Each of the duplexes involved in these junctions has unique sequences at its ends, which allowed the oligonucleotides to anneal to form the indicated Holliday junctions; they also possess a core of 11 nucleotides that are homologous. he Holliday junction
can branch-migrate within the core region. A dimer of RuvC binds to
Holliday junctions and cleaves a pair of adjacent strands between nucleotides 3 and 4 in the sequences 5ʹ-ATTG, 5ʹ-ATTC, 5ʹ-TTTG, or 5ʹ-TTTC,
as shown for 5ʹ-ATTG in Figure 5–31A. hese four sequences are represented by the consensus sequence 5ʹ-A/TTTG/C.
You want to know whether the two subunits of the RuvC dimer coordinate the cleavages on the two strands or can act independently. To investigate this you make the three Holliday junctions shown in Figure 5–31A:
one with two cleavable sequences, one with two uncleavable sequences,
and one with one cleavable and one uncleavable sequence. You label
the 5ʹ end of one strand in each, incubate them with RuvC, and analyze
cleavage by electrophoresis on denaturing gels, so that the oligonucleotides separate from one another. he results are shown in Figure 5–31B.
A. What fraction of all possible four-nucleotide sequences is cleaved by
RuvC?
B. Does the requirement for resolution at a limited set of speciic sequences
seriously restrict the sites in the E. coli genome at which Holliday junctions can be resolved?
C. Do the two subunits of the RuvC dimer coordinate their cleavage of the
two strands or can they act independently? Explain your reasoning.
D. Draw out the duplex products generated by RuvC resolution of the “consensus” Holliday junction in Figure 5–31A, the one with its ends marked
by letters.
TRANSPOSITION AND CONSERVATIVE SITE-SPECIFIC
RECOMBINATION
TERMS TO LEARN
conservative site-specific
recombination
DNA-only transposon
nonretroviral
retrotransposon
phase variation
retroviral-like
retrotransposon
retrovirus
reverse transcriptase
transposable element
transposition
transposon
99
Figure 5–31 Analysis of Holliday
junction cleavage (Problem 5–84).
(A) Effects of DNA sequence on
resolution of Holliday junctions. Cleavable
sequences are shown in yellow boxes;
uncleavable sequences are shown in
blue boxes. Arrows indicate the sites
of cleavage in a Holliday junction with
consensus sequences. (B) Resolution of
Holliday junctions by RuvC. Increasing
concentrations of RuvC in each lane
are indicated schematically by triangles.
The locations of the labeled 5ʹ ends are
indicated by black dots.
100
Chapter 5: DNA Replication, Repair, and Recombination
DEFINITIONS
(A)
BREAK
Match each deinition below with its term from the list above.
5–85
Length of DNA that moves from a donor site to a target site either by cutand-paste transposition or by replicative transposition.
5–86
Enzyme that makes a double-strand DNA copy from a single-strand RNA
template molecule.
5–87
RNA-containing virus that replicates in a cell by irst making a doublestrand DNA intermediate.
5–88
Rearrangement of DNA that depends on the breakage and rejoining of
two DNA helices at speciic sequences on each DNA molecule.
TRUE/FALSE
Consider the following statement and explain your answer.
5–89
When transposable elements move around the genome, they rarely
integrate into the middle of a gene because gene disruption—a potentially lethal event to the cell and the transposon—is selected against by
evolution.
REJOIN
(B)
a
b
c
d
a
b
c
d
Figure 5–32 Cre recombinase-mediated
site-speciic recombination (Problem
5–90). (A) Schematic representation of
Cre/loxP site-speciic recombination.
The loxP sequences in the DNA are
represented by triangles that are colored
so that the site-speciic recombination
event can be followed more readily. In
reality their DNA sequences are identical.
(B) DNA substrates containing two
arrangements of loxP sites.
THOUGHT PROBLEMS
5–90
Cre recombinase is a site-speciic enzyme that catalyzes recombination
between two LoxP DNA recognition sequences. Cre recombinase pairs
two LoxP sites in the same orientation, breaks both duplexes at the same
point in each LoxP site, and joins the ends with new partners so that each
LoxP site is regenerated, as shown schematically in Figure 5–32A. Based
on this mechanism, predict the arrangement of sequences that will be
generated by Cre-mediated site-speciic recombination for each of the
two DNAs shown in Figure 5–32B.
DATA HANDLING
5–91
You are studying the prokaryotic transposon Tn10 and have just igured
out an elegant way to determine whether Tn10 replicates during transposition or moves by a cut-and-paste mechanism. Your idea is based on the
key diference between these two mechanisms: both parental strands of
the Tn10 move in cut-and-paste transposition, whereas only one parental strand moves in replicative transposition (Figure 5–33).
You plan to mark the individual strands by annealing strands from
two diferent Tn10s. Both Tn10s contain a gene for tetracycline resistance and a gene for lactose metabolism (LacZ), but in one, the LacZ gene
is inactivated by a mutation. his diference provides a convenient way
to follow the two Tn10s since LacZ+ bacterial colonies (when incubated
with an appropriate substrate) turn blue, but LacZ– colonies remain
white. You denature and reanneal a mixture of the two transposon DNAs,
which produces an equal mixture of heteroduplexes and homoduplexes
λ DNA
Tn10
bacterial genome
cut-and-paste
transposition
replicative
transposition
Figure 5–33 Replicative and cut-andpaste transposition of a transposable
element (Problem 5–91). The transposable
element is shown as a heteroduplex,
which is composed of two genetically
different strands—one red and one
yellow. During replicative transposition,
one strand stays with the donor DNA and
one strand is transferred to the bacterial
genome. In cut-and-paste transposition,
the transposable element is cut out of the
donor DNA and transferred entirely to the
bacterial genome.
TRANSPOSITION AND CONSERVATIVE SITE-SPECIFIC RECOMBINATION
(Figure 5–34). You introduce them into LacZ– bacteria, and spread the
bacteria onto Petri dishes that contain tetracycline and the color-generating substrate.
Once inside a bacterium, the transposon will move (at very low frequency) into the bacterial genome, where it confers tetracycline resistance on the bacterium. he rare bacterium that gains a Tn10 survives the
selective conditions and forms a colony. When you score a large number
of such colonies, you ind that roughly 25% are white, 25% are blue, and
50% are mixed, with one blue sector and one white sector.
A. Explain the source of each kind of bacterial colony and decide whether
the results support a replicative or a cut-and-paste mechanism for Tn10
transposition.
B. You performed these experiments using a recipient strain of bacteria that
was incapable of repairing mismatches in DNA. How would you expect
the results to difer if you used a bacterial strain that could repair mismatches?
5–92
he Ty elements of the yeast Saccharomyces cerevisiae move to new locations in the genome by transposition through an RNA intermediate.
Normally, the Ty-encoded reverse transcriptase is expressed at such a
low level that transposition is very rare. To study the transposition process, you engineer a cloned version of the Ty element so that the gene for
reverse transcriptase is linked to the galactose control elements. You also
“mark” the element with a segment of bacterial DNA so that you can
detect it speciically and thus distinguish it from other Ty elements in the
genome. As a target gene to detect transposition, you use a defective histidine (His–) gene whose expression is dependent on the insertion of a
Ty element near its 5ʹ end. You show that yeast cells carrying a plasmid
with your modiied Ty element generate His+ colonies at a frequency of
5 × 10–8 when grown on glucose. When the same cells are grown on galactose, the frequency of His+ colonies is 10–6, a 20-fold increase.
You notice that cultures of cells with the Ty-bearing plasmid grow
normally on glucose but very slowly on galactose. To investigate this phenomenon, you isolate individual colonies that arise under three diferent
conditions: His– colonies grown in the presence of glucose, His– colonies
grown in the presence of galactose, and His+ colonies grown in the presence of galactose. You eliminate the plasmid from each colony, isolate
DNA from each culture, and analyze it by gel electrophoresis and blot
hybridization, using the bacterial marker DNA as a probe. Your results
are shown in Figure 5–35.
A. Why does transposition occur so much more frequently in cells grown on
galactose than it does in cells grown on glucose?
B. As shown in Figure 5–35, His– cells isolated after growth on galactose have
about the same number of marked Ty elements in their chromosomes as
His+ cells that were isolated after growth on galactose. If transposition is
independent of histidine selection, why is the frequency of Ty-induced
His+ colonies so low (10–6)?
C. Why do you think it is that cells with the Ty-bearing plasmid grow so
slowly on galactose?
glucose (His–)
1
2
3
4
galactose (His–)
5
6
7
8
101
λ
λ
Tn10
LacZ +
Tn10
LacZ
DENATURE
REANNEAL
heteroduplex
homoduplex
Figure 5–34 Formation of a mixture of
heteroduplexes and homoduplexes by
denaturing and reannealing two
different Tn10 genomes (Problem 5–91).
The Tn10 DNA is shown as a box. The
sequence differences between the two
Tn10s are indicated by the blue and
yellow segments.
galactose (His+)
9 10 11 12 13 14 15
Figure 5–35 Analysis of cells that
harbored a plasmid carrying a modiied
Ty element (Problem 5–92). Cells were
initially grown on glucose or galactose as
a carbon source. His– cells were grown in
the presence of histidine; His+ cells were
grown in its absence. Bands indicate
restriction fragments that hybridized to
the marker DNA originally present in the
Ty element carried by the plasmid.
102
Chapter 5: DNA Replication, Repair, and Recombination
MCAT STYLE
Passage 1 (Questions 5–93 to 5–95)
Taq DNA polymerase, which comes from the thermotolerant bacterium hermus
aquaticus, is used extensively for the polymerase chain reaction (PCR) because it
remains functional at the high temperatures required for PCR. In a typical PCR,
a DNA sample is heated to separate the two strands; two short single strands of
DNA, called primers, are annealed to their complementary sequences in the DNA
on either side of the segment to be ampliied; and Taq polymerase then synthesizes DNA starting from the primers. After synthesis has occurred, the reaction
is heated to high temperatures to separate the DNA strands, and then cooled to
begin another round of primer-directed DNA synthesis. he cycle is repeated
many times, leading to ampliication of the DNA between the primers.
5–93
One problem with Taq polymerase is that it incorporates the wrong base
approximately once every 8000 bases, a frequency of mistakes that is far
higher than the error rates for DNA polymerases that carry out DNA replication in cells. Which one of the following statements best explains the
high error rate of Taq polymerase in PCR?
A. Taq polymerase cannot make Okazaki fragments eiciently.
B. Taq polymerase lacks a 3ʹ-to-5ʹ proofreading exonuclease.
C. Taq polymerase often falls of the DNA, interrupting synthesis.
D. Taq polymerase synthesizes DNA in the 3ʹ-to-5ʹ direction.
5–94
Another problem with Taq polymerase is that it is only capable of synthesizing relatively small pieces of DNA, with an upper limit of around 4000
bases. Which one of the following is a likely explanation for the inability
of Taq polymerase to synthesize long pieces of DNA?
A. Taq polymerase can make only relatively short Okazaki fragments.
B. Taq polymerase cannot remove mismatched bases from the 3ʹ end.
C. Taq polymerase lacks the helicase required for strand separation.
D. Taq polymerase reactions lack the primase needed for new primers.
5–95
Imagine that you have used PCR to amplify a speciic region of DNA about
4000 bases long. You then ligate the PCR products into a plasmid vector
and transform it into bacteria. One of the transformed bacteria receives
a plasmid that includes a PCR product with two separate mismatched
bases resulting from Taq polymerase errors the last time the DNA was
copied. You grow this bacterium into a colony and isolate the population
of plasmids derived from the single transformed plasmid. Assuming that
the mismatches were resolved by the cell’s mismatch repair machinery
prior to replication of the plasmid, what are the chances that both mismatches were converted to mutations?
A. 100% chance
B. 50% chance
C. 25% chance
D. 0% chance
Passage 2 (Question 5–96)
he irst DNA polymerase to be discovered was puriied from E. coli and called
DNA polymerase I. It synthesized DNA at a rate of about 10–20 bases per second,
but could only synthesize about 25–30 bases before falling of the template DNA.
In addition to its polymerase activity, it also possessed both 5ʹ-to-3ʹ and 3ʹ-to-5ʹ
exonuclease activities, which could act on DNA or RNA hybridized to a DNA template. Another DNA polymerase—DNA polymerase III—was discovered later.
It could synthesize much longer stretches of DNA at a rate of about a thousand
bases per second, and had only a 3ʹ-to-5ʹ exonuclease activity.
MCAT STYLE
5–96
Based on this information, which one of the following functions best
describes the role of DNA polymerase I in bacterial cells?
A. Modiies Okazaki fragments for joining into intact strands.
B. Repairs any DNA that is damaged during transcription.
C. Synthesizes most of the cellular DNA during replication.
D. Synthesizes RNA primers to initiate Okazaki fragments.
103
Reading the Genome.
The complete sequence of the human
genome printed out occupies 7 shelves
of 16 volumes, each containing hundreds
of pages like the one shown here, with
almost 10 million characters in each
volume printed in 4-point courier type
without any index. Try to see if you can
ind the gene in this sequence, which
comes from human chromosome 11,
open on the shelf in the photograph that
was taken at the opening of the Wellcome
Museum, london, in July 2007.
Chapter 6
105
How Cells Read the Genome:
From DNA to Protein
CHAPTER
6
FROM DNA TO RNA
IN THIS CHAPTER
TERMS TO LEARN
FROM DNA TO RNA
consensus nucleotide sequence
DNA supercoiling
DNA transcription
exon
exosome
general transcription factor
intron
mRNA (messenger RNA)
noncoding RNA
nuclear pore complex
promoter
RNA polymerase
RNA splicing
rRNA gene
snoRNA (small nucleolar RNA)
snRNA (small nuclear RNA)
spliceosome
TATA box
terminator
DEFINITIONS
Match the deinition below with its term from the list above.
6–1
Helps to position the RNA polymerase correctly at the promoter, to aid in
pulling apart the two strands of DNA to allow transcription to begin, and
to release RNA polymerase from the promoter into the elongation mode
once transcription has begun.
6–2
Small RNA molecules that are complexed with proteins to form the ribonucleoprotein particles involved in RNA splicing.
6–3
Nucleotide sequence in DNA to which RNA polymerase binds to begin
transcription.
6–4
A large protein complex containing multiple 3ʹ-to-5ʹ RNA exonucleases
that degrade improperly processed mRNAs, introns, and other RNA
debris retained in the nucleus.
6–5
he enzyme that carries out transcription.
6–6
RNA molecule that speciies the amino acid sequence of a protein.
6–7
Process in which intron sequences are excised from RNA transcripts in
the nucleus during the formation of messenger and other RNAs.
6–8
Signal in bacterial DNA that halts transcription.
6–9
Segment of a eukaryotic gene consisting of a sequence of nucleotides
that will be represented in mRNA or other functional RNAs.
6–10
Large multiprotein structure forming a channel through the nuclear
envelope that allows selected molecules to move between nucleus and
cytoplasm.
FROM RNA TO PROTEIN
THE RNA WORLD AND THE
ORIGINS OF LIFE
106
Chapter 6: How Cells Read the Genome: From DNA to Protein
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
6–11
he consequences of errors in transcription are less severe than those of
errors in DNA replication.
6–12
he σ subunit is a permanent component of the RNA polymerase holoenzyme from E. coli, allowing it to initiate at appropriate promoters in the
bacterial genome.
6–13
Eukaryotic mRNA molecules carry 3ʹ ribosyl OH groups at both their 3ʹ
and 5ʹ ends.
6–14
Since introns are largely genetic “junk,” they do not have to be removed
precisely from the primary transcript during RNA splicing.
6–15
RNA polymerase II generates the end of a pre-mRNA transcript when
it ceases transcription and releases the transcript; a poly-A tail is then
quickly added to the free 3ʹ end.
THOUGHT PROBLEMS
6–16
Consider the expression “central dogma,” which refers to the proposition
that genetic information lows from DNA to RNA to protein. Is the word
“dogma” appropriate in this scientiic context?
6–17
In the electron micrograph in Figure 6–1, are the RNA polymerase molecules moving from right to left or from left to right? How can you tell?
Why do the RNA transcripts appear so much shorter than the length of
DNA that encodes them?
6–18
6–19
Match the following list of RNAs with their functions.
A. mRNA
1. blocks translation of selected mRNAs
B. rRNA
2. modiication and processing of rRNA
C. snoRNA
3. protects germ line from transposable elements
D. snRNA
4. components of ribosome
E. tRNA
5. splicing of RNA transcripts
F. piRNA
6. directs degradation of selected mRNAs
G. miRNA
7. codes for proteins
H. siRNA
8. adaptor for protein synthesis
An RNA polymerase is transcribing a segment of DNA that contains the
sequence
5ʹ-GTAACGGATG-3ʹ
3ʹ-CATTGCCTAC-5ʹ
If the polymerase transcribes this sequence from left to right, what will
the sequence of the RNA be? What will the RNA sequence be if the
polymerase moves right to left?
6–20
What are the roles of general transcription factors in RNA polymerase IImediated transcription, and why are they referred to as “general”?
Figure 6–1 Transcription of two adjacent
rRNA genes (Problem 6–17). The scale bar
is 1 μm. (Courtesy of Ulrich Scheer.)
FROM DNA TO RNA
107
magnet
+
+
–
–
fluorescent
beads
magnetic
bead
Figure 6–2 Supercoils around a moving RNA polymerase (Problem 6–21).
DNA
6–21
In which direction along the template must the RNA polymerase in
Figure 6–2 be moving to have generated the supercoiled structures that
are shown? Would you expect supercoils to be generated if the RNA polymerase were free to rotate about the axis of the DNA as it progressed
along the template?
6–22
You have attached an RNA polymerase molecule to a glass slide and have
allowed it to initiate transcription on a template DNA that is tethered to
a magnetic bead as shown in Figure 6–3. If the DNA with its attached
magnetic bead moves relative to the RNA polymerase as indicated in the
igure, in which direction will the bead rotate?
6–23
Why doesn’t transcription cause a hopeless tangle? If the RNA polymerase does not revolve around the DNA as it moves, it will induce two
DNA supercoils—one in front and one behind—for every 10 nucleotides
it transcribes. If, instead, RNA polymerase revolves around the DNA—
avoiding DNA supercoiling—then it will coil the RNA around the DNA
duplex, once for every 10 nucleotides it transcribes. hus, for any reasonable-size gene, the act of transcription should result in hundreds of
coils or supercoils…and that’s for every single RNA polymerase! So why
doesn’t transcription lead to a complete snarl?
6–24
You are studying a DNA virus that makes a set of abundant proteins late
in its infectious cycle. An mRNA for one of these proteins maps to a DNA
restriction fragment from the middle of the linear genome. To determine
the precise location of this mRNA you anneal it with the puriied restriction fragment under conditions where only DNA–RNA hybrid duplexes
are stable and DNA strands do not reanneal. When you examine the
reannealed DNA–RNA duplexes by electron microscopy, you see structures such as that in Figure 6–4. Why are there single-stranded tails at the
ends of the DNA–RNA duplex region?
6–25
Smilin is a (hypothetical) protein that causes people to be happy. It is
inactive in many chronically unhappy people. he mRNA isolated from
a number of diferent unhappy individuals in the same family was found
to lack an internal stretch of 173 nucleotides that are present in the Smilin mRNA isolated from a control group of generally happy people. he
DNA sequences of the Smilin genes from the happy and unhappy persons were determined and compared. hey difered by just one nucleotide change—and no nucleotides were deleted. Moreover, the change
was found in an intron.
A. Can you hypothesize a molecular mechanism by which a single nucleotide change in a gene could cause the observed internal deletion in the
mRNA?
B. What consequences for the Smilin protein would result from removing
a 173-nucleotide-long internal stretch from its mRNA? Assume that the
173 nucleotides are deleted from the coding region of the Smilin mRNA.
C. What can you say about the molecular basis of unhappiness in this
family?
6–26
he human α-tropomyosin gene is alternatively spliced to produce different forms of α-tropomyosin mRNA in diferent cell types (Figure 6–5).
RNA
RNA
polymerase
glass slide
Figure 6–3 System for measuring
the rotation of DNA caused by RNA
polymerase (Problem 6–22). The magnet
holds the bead upright (but doesn’t
interfere with its rotation), and the
attached tiny luorescent beads allow the
direction of motion to be visualized under
the microscope. rna polymerase is held
in place by attachment to the glass slide.
Figure 6–4 Dna–rna hybrid between an
mrna and a Dna restriction fragment
from adenovirus (Problem 6–24).
Chapter 6: How Cells Read the Genome: From DNA to Protein
108
(A) HUMAN α-TROPOMYOSIN GENE
1
4
5
23
6
910
78
11
12
13
(B) FOUR DIFFERENT SPLICE VARIANTS
Figure 6–5 Alternatively spliced mRNAs
from the human α-tropomyosin gene
(Problem 6–26). (A) Exons in the human
α-tropomyosin gene. The locations and
relative sizes of exons are shown by
blue and red rectangles, with alternative
exons in red. (B) Splicing patterns for
four α-tropomyosin mRNAs. Splicing is
indicated by lines connecting the exons
that are included in the mRNA.
For all forms of the mRNA, the protein sequences encoded by exon 1 are
the same, as are the protein sequences encoded by exon 10. Exons 2 and
3 are alternative exons used in diferent mRNAs, as are exons 7 and 8.
Which of the following statements about exons 2 and 3 is the most accurate? Is that statement also the most accurate one for exons 7 and 8?
Explain your answers.
A. Exons 2 and 3 must have the same number of nucleotides.
B. Exons 2 and 3 must each contain an integral number of codons (that is,
the number of nucleotides divided by 3 must be an integer).
C. Exons 2 and 3 must each contain a number of nucleotides that when
divided by 3 leaves the same remainder (that is, 0, 1, or 2).
6–27
You have printed out a set of DNA sequences around the intron/exon
boundaries for genes in the β-globin family, and have taken the thick ile
to the country to study for the weekend. When you look at the printout,
you discover to your annoyance that there’s no indication of where in the
gene you are. You know that the sequences in Figure 6–6 come from one
of the exon/intron or intron/exon boundaries and that the boundaries lie
on the dotted line, but you don’t know the order of the intron and exon.
You know that introns begin with the dinucleotide sequence GT and end
with AG, but you realize that these particular sequences would it either
as the start or the inish of an intron.
If you cannot decide which side is the intron, you will have to cut your
weekend short and return to the city (or ind a neighbor with Internet
access). In desperation, you consider the problem from an evolutionary
perspective. You know that introns evolve faster (sufering more nucleotide changes) than exons because they are not constrained by function.
Does this perspective allow you to identify the intron, or will you have to
pack your bags?
exon 1
intron
GT
exon 2
AG
exon 1
intron
GT
GGTGGTGAGGCCCTGGGCAG GTAGGTATCCCACTTACAAG - cow
GGTGGTGAGATTCTGGGCAG GTAGGTACTGGAAGCCGGGG - gorilla
GGGGGCGAAGCCCTGGGCAG GTAGGTCCAGCTTCGGCCAT - chicken
GGTGGTGAGGCCCTGGGCAG GTTGGTATCAAGGTTACAAG - human
GGTGGTGAGGCCCTGGGCAG GTTGGTATCCAGGTTACAAG - mouse
GGTGGTGAGGCCCTGGGCAG GTTGGTATCCTTTTTACAGC - rabbit
GGCCATGATGCCCTGACCAG GTAACTTGAAGCACATTGCT - frog
AG
intron
exon 2
Figure 6–6 Aligned DNA sequences from
the β-globin genes in different species
(Problem 6–27). As indicated by the gene
structures shown above and below, the
DNA sequences could come from the
boundary of exon 1 with the intron, or from
the boundary of the intron with exon 2.
FROM DNA TO RNA
Figure 6–7 A test for intron scanning during RNA splicing (Problem 6–28).
(A) Minigene with two 3ʹ splice sites. (B) Minigene with two 5ʹ splice sites.
Boxes represent complete (rectangles) or partial (ragged edge) exons;
5ʹ and 3ʹ splice junctions are indicated.
6–28
Many eukaryotic genes contain a large number of exons. Correct splicing
of such genes requires that neighboring exons be ligated to one another;
if they are not, exons will be left out. Since 5ʹ splice sites look alike, as
do 3ʹ splice sites, it is remarkable that skipping an exon occurs so rarely.
Some mechanism must keep track of neighboring exons and ensure that
they are brought together.
One early proposal suggested that the splicing machinery bound to a
splice site at one end of an intron and scanned through the intron to ind
the splice site at the other end. Such a scanning mechanism would guarantee that an exon was never skipped. his hypothesis was tested with
two minigenes: one with a duplicated 5ʹ splice site and the other with a
duplicated 3ʹ splice site (Figure 6–7). hese minigenes were transfected
into cells and their RNA products were analyzed to see which 5ʹ and 3ʹ
splice sites were selected during splicing.
A. Draw a diagram of the products you expect from each minigene if the
splicing machinery binds to a 5ʹ splice site and scans toward a 3ʹ splice
site. Diagram the expected products if the splicing machinery scans in
the opposite direction.
B. When the RNA products from the transfected minigenes were analyzed,
it was found that each minigene generated a mixture of the two possible
5ʹ-to-3ʹ splice products. Based on these results, would you conclude that
neighboring exons are brought together by intron scanning? Why or why
not?
6–29
What does “export ready” mRNA mean, and what distinguishes an “export
ready” mRNA from a bit of excised intron that needs to be degraded?
6–30
he nucleolus disappears at each mitosis and then reappears during G1
of the next cell cycle. How is this reversible process thought to be accomplished?
CALCULATIONS
6–31
You have established a transcription assay in which transcripts initiate
at a speciic adenovirus promoter in a plasmid (see Problem 6–35). Each
transcript is 400 nucleotides long and has an overall composition of C2AU.
hese transcripts accumulate linearly for about an hour and then reach
a plateau. Your assay conditions use a 25 μL reaction volume containing
16 μg/mL of DNA template (the plasmid, which is 3.5 kb in length) with
all other components in excess. From the speciic activity of the 32P-CTP
and the total radioactivity in the transcripts, you calculate that at the
plateau 2.4 pmol of CMP was incorporated. (he mass of a nucleotide
pair is 660 daltons.)
A. How many transcripts are produced per reaction?
B. How many templates are present in each reaction?
C. How many transcripts are made per template in the reaction?
DATA HANDLING
6–32
If RNA polymerase proofreads its product in a manner analogous to DNA
polymerase, it must slow its rate of nucleotide incorporation after adding
an incorrect base to the end of the growing RNA chain. his delay will
allow time for removal of the mismatched nucleotide before the next one
is added.
109
(A) MINIGENE 1
5′
3′
3′
(B) MINIGENE 2
5′
5′
3′
110
Chapter 6: How Cells Read the Genome: From DNA to Protein
TABLE 6–1 Rate of incorporation of correct and incorrect nucleotides
(Problem 6–32).
Incorporated nucleotide (bold)
(A) TEMPLATE FOR TRANSCRIPTION
Vmax (units/sec)
CUA
0.20
CUG
0.0015
CUAC
0.17
CUGC
0.036
3‘
RNA --CUUAUCCUCU
DNA --GAATAGGAGATGTCA
34
43
template strand
5‘
(B) EXPERIMENTAL DATA
To investigate this possibility, scientists devised a clever technique
to measure the rate of nucleotide incorporation by RNA polymerase at
a deined point in a DNA template. he template contained a promoter
and was covalently attached to agarose beads, so that it (and the attached
RNA polymerase) could be removed from solution and washed, and then
reincubated in a new mixture of nucleotides (Figure 6–8A). By using a
series of solutions containing just one or a couple of nucleotides, the
RNA polymerase was “walked” along the template to the site shown in
Figure 6–8A. At that point, the RNA polymerase and template were resuspended in a solution containing one nucleotide, and the rate of incorporation of that nucleotide was measured as shown in Figure 6–8B. he
rates of incorporation of A and G at position +44 in the RNA strand are
shown in Table 6–1. In addition, the incorporation of the following C at
position +45 was measured after A or G was incorporated at position +44
(Table 6–1).
A. Imagine that the RNA polymerase was stopped so that the last nucleotide
in the RNA chain was the C at position +34. Describe how you might
“walk” the polymerase from there to position +43 to do these experiments.
B. Is the correct nucleotide preferred over the incorrect nucleotide? If so,
by what factor is it preferred? Does the presence of an incorrect nucleotide inluence the rate of addition of the next nucleotide? Explain your
answer.
6–33
In Figure 6–9, the sequences of 13 promoters recognized by the σ70 factor
of RNA polymerase have been aligned. Deduce the consensus sequences
for the –10 and –35 regions of these promoters.
6–34
Deletion analysis of protein-binding sequences in a promoter can be dificult to interpret because altered spacing between elements can critically afect their function. he “linker-scanning” method eliminates this
potential diiculty by replacing 10-nucleotide segments throughout the
promoter with oligonucleotide linkers. A classic paper described this
method in an analysis of the promoter for the thymidine kinase (Tk)
time after adding ATP (seconds)
0
43
4
6
10
30
60
180
44
Figure 6–8 Determination of kinetic
parameters for RNA polymerase (Problem
6–32). (A) DNA template attached to an
agarose bead. (B) Rate of incorporation
of the correct A nucleotide in the nascent
RNA chain. (The bands in adjacent lanes
form a curve because of an artifact of gel
electrophoresis.)
+1
tyrosine tRNA
promoter
ribosomal RNA
gene promoters
bacteriophage
promoters
TCTCAACGTAACACTTTACAGCGGCG--CGTCATTTGATATGATGC-GCCCCGCTTCCCGATAAGGG
GATCAAAAAAATACTTGTGCAAAAAA--TTGGGATCCCTATAATGCGCCTCCGTTGAGACGACAACG
ATGCATTTTTCCGCTTGTCTTCCTGA--GCCGACTCCCTATAATGCGCCTCCATCGACACGGCGGAT
CCTGAAATTCAGGGTTGACTCTGAAA--GAGGAAAGCGTAATATAC-GCCACCTCGCGACAGTGAGC
CTGCAATTTTTCTATTGCGGCCTGCG--GAGAACTCCCTATAATGCGCCTCCATCGACACGGCGGAT
TTTTAAATTTCCTCTTGTCAGGCCGG--AATAACTCCCTATAATGCGCCACCACTGACACGGAACAA
GCAAAAATAAATGCTTGACTCTGTAG--CGGGAAGGCGTATTATGC-ACACCCCGCGCCGCTGAGAA
TAACACCGTGCGTGTTGACTATTTTA-CCTCTGGCGGTGATAATGG--TTGCATGTACTAAGGAGGT
TATCTCTGGCGGTGTTGACATAAATA-CCACTGGCGGTGATACTGA--GCACATCAGCAGGACGCAC
GTGAAACAAAACGGTTGACAACATGA-AGTAAACACGGTACGATGT-ACCACATGAAACGACAGTGA
TATCAAAAAGAGTATTGACTTAAAGT-CTAACCTATAGGATACTTA-CAGCCATCGAGAGGGACACG
ACGAAAAACAGGTATTGACAACATGAAGTAACATGCAGTAAGATAC-AAATCGCTAGGTAACACTAG
GATACAAATCTCCGTTGTACTTTGTT--TCGCGCTTGGTATAATCG-CTGGGGGTCAAAGATGAGTG
Figure 6–9 Sequences
recognized by σ70 factor
(Problem 6–33). Different
σ factors are designated by their
molecular masses; σ70 has a
mass of 70 kilodaltons. Dashes
represent spaces that have been
added to maximize alignment of
sequences in the –10 and
–35 regions.
FROM DNA TO RNA
111
positions of linker relative to transcription start site
–120
–100
–80
–60
–40
–20
+1
+20
10
linker-scanned
transcript
control
transcript
gene. Plasmids in which 10-nucleotide segments had been replaced
with linkers were injected into Xenopus laevis oocytes along with control
plasmids to measure injection eiciency. Results of these injections are
shown in Figure 6–10.
A. Estimate from these experiments the locations of sequences that are critical for promoter function, and rank their relative importance.
B. Which, if any, of these elements do you suppose corresponds to the TATA
box?
6–35
Figure 6–10 linker-scanning analysis
of the Tk promoter (Problem 6–34).
Transcripts from linker-scanned plasmids
and control plasmids were analyzed by
primer extension, using a radiolabeled
primer corresponding to sequences
about 80 nucleotides from the 5ʹ end
of the transcript. These primers were
extended to the 5ʹ ends of the transcripts
and the products were displayed on
a denaturing polyacrylamide gel. Two
bands are present for both the control
and linker-scanned transcripts because of
ineficient extension to the very end of the
transcript. The position of each linker is
indicated at the center of the segment it
replaced. a 10-base-pair bar—the length
of the replacements—is shown at the top
left of the igure.
Puriication of a transcription factor typically requires a rapid assay to
prevent inactivation of the factor before it can be identiied. One key
technical advance was to use a promoter linked to a 400-nucleotide DNA
sequence that contained no C nucleotides. When GTP is omitted from
the assay mixture (but CTP, UTP, and ATP are included), the only long
RNA transcript is made from the synthetic DNA sequence because all
other transcripts terminate when a G is required. his set-up allows a
rapid assay of speciic transcription simply by measuring the incorporation of a radioactive nucleotide into the long transcript.
To test this idea, two plasmids were constructed that carried the synthetic sequence: one with a promoter from adenovirus (pML1), the other
without a promoter (pC1) (Figure 6–11A). Each plasmid was mixed with
pure RNA polymerase II, transcription factors, UTP, ATP, and 32P-CTP.
In addition, various combinations of GTP, RNAse T1 (which cleaves
RNA adjacent to each G nucleotide), and 3ʹ O-methyl GTP (which terminates transcription whenever G is incorporated) were added. he
products were measured by gel electrophoresis, with the results shown
in Figure 6–11B.
A. Why is the 400-nucleotide transcript absent from lane 4 but present in
lanes 2, 6, and 8?
(B) IN VITRO TRANSCRIPTION ASSAYS
(A) TEST PLASMIDS
adenovirus
promoter
synthetic
insert
(400 bp)
pML1
plasmid
C
ML
C
ML
C
ML
C
ML
GTP
–
–
+
+
+
+
+
+
RNase T1
–
–
–
–
+
+
+
+
3’ O-methyl GTP
–
–
–
–
–
–
+
+
vector
synthetic
insert
(400 bp)
400
nucleotides
pC1
vector
1
2
3
4
5
6
7
8
Figure 6–11 Characterization of
transcription using a template without
C nucleotides (Problem 6–35).
(a) structures of test plasmids. (B) results
of transcription assays under various
conditions. all reactions contain rna
polymerase II, transcription factors, UTp,
aTp, and 32p-CTp. Other components are
listed above each lane. C is plasmid pC1;
ML is plasmid pML1.
112
Chapter 6: How Cells Read the Genome: From DNA to Protein
(A) COLUMN CHROMATOGRAPHY
1
(B) DEADLY MUSHROOMS
protein concentration
RNA polymerase activity
no α -amanitin
1 µg/mL α -amanitin
10 µg/mL α -amanitin
protein
trace
2
3
Amanita phalloides
10
20
30
40
50
fraction number
B. Can you guess the source of the synthesis in lane 3 when the promoterless pC1 plasmid is used?
C. Why is a transcript of about 400 nucleotides present in lane 5 but not in
lane 7?
D. he goal in developing this ingenious assay was to aid the puriication
of transcription factors; however, that process will begin with crude cell
extracts, which will contain GTP. How do you suppose you might assay
speciic transcription in crude extracts?
6–36
When RNA polymerases were irst being characterized in eukaryotes,
three peaks of polymerizing activity (1, 2, and 3 in Figure 6–12A) were
commonly obtained by fractionating cell extracts on chromatography
columns. It was unclear whether these peaks corresponded to diferent
RNA polymerases or just to diferent forms of one polymerase. Incubating
the three polymerase fractions in the presence of 1 μg/mL or 10 μg/mL
α-amanitin (from Amanita phalloides, the world’s deadliest mushroom;
Figure 6–12B) gave the results shown in Figure 6–12A. Do these results
argue for diferent RNA polymerases or diferent forms of the same RNA
polymerase? Explain your answer.
6–37
he large subunit of eukaryotic RNA polymerase II in yeast has a CTD
(C-terminal domain) that comprises 27 near-perfect repeats of the
sequence YSPTSPS. If the normal RNA polymerase II gene is replaced
with one that encodes a CTD with only 11 repeats, the cells are viable at
30°C but are unable to grow at 12°C. his cold sensitivity allows suppressors to be selected for growth at 12°C. Some of these suppressors proved
to be dominant mutations in the previously unknown gene Srb2.
Extracts prepared from yeast that are lacking the Srb2 gene cannot
transcribe added DNA templates, but they can be activated for transcription by the addition of Srb2 protein. To test the role of Srb2 in transcription, plasmid DNAs with either a short or a long G-free sequence downstream of a promoter (Figure 6–13A) were incubated separately in the
presence or absence of a limiting amount of Srb2 [and in the absence
of added nucleotide triphosphates (NTPs) so that transcription could
not begin]. he reactions were then mixed and transcription was initiated at various times afterward by adding a mixture of all four NTPs that
included 32P-CTP (Figure 6–13B). After a brief incubation (so that transcription did not have time to reinitiate) the products were displayed by
gel electrophoresis. Whichever template was preincubated with Srb2 was
the one that was transcribed (Figure 6–13C). By contrast, if an excess of
Srb2 was mixed with one template during the preincubation, transcription was observed from both templates after mixing.
Figure 6–12 Characterization
of RNA polymerase activities
(Problem 6–36). (A) Peaks of
RNA polymerase activity after
column chromatography.
Activities were measured
in the presence of various
concentrations of α-amanitin.
(B) The Amanita phalloides
mushroom. (Courtesy of Fred
Stevens.)
FROM DNA TO RNA
113
(A) TEMPLATES
Figure 6–13 Experiments to test the role
of Srb2 in transcription (Problem 6–37).
(A) Short (S) and long (l) G-free
templates. (B) Experimental design.
(C) Experimental results. The transcripts
from the S and l templates were
visualized by autoradiography.
promoter
L
promoter
S
(B) DESIGN OF THE EXPERIMENT
MIX
PREINCUBATE
ASSAY
L or S template, + Srb2
take samples,
assay transcription
S or L template, – Srb2
0
60
70
80
90
time (minutes)
(C) EXPERIMENTAL RESULTS
TEMPLATES
preincubation + Srb2 L
S
L+S
–
L
L
L
L
S
S
S
preincubation – Srb2
–
–
–
–
S
S
S
S
L
L
L
L
time of assay (min)
60
60
60
60
60
70
80
90
60
70
80
90
1
2
3
4
5
6
7
8
9
10
11
12
S
L transcript
S transcript
A. Did Srb2 show a preference for either template when the preincubation was carried out with the individual templates or the mixture (Figure
6–13C, lanes 1 to 3)?
B. Do the results indicate that Srb2 acts stoichiometrically or catalytically?
How so?
C. Is Srb2 part of the complex of proteins that forms on the template before
transcription is initiated, or does it act after transcription has begun?
How can you tell?
D. What do you think happens during the preincubation that so strongly
favors transcription from the template that was included in the preincubation?
E. Do these results indicate that Srb2 binds to the CTD of RNA polymerase
II?
6–38
Detailed features of the active site of RNA polymerase have been probed
using modiied RNA nucleotides that can react with nearby targets. In one
study, an analog of U carrying a reactive chemical group (Figure 6–14A)
was incorporated at one of two positions in a radiolabeled RNA and then
“walked” to speciic locations relative to the 3ʹ end of the RNA chain using
a technique like that described in Problem 6–32 (Figure 6–14B). When
the analog was then activated, it reacted almost exclusively either with
the adenine it was paired with in the DNA template strand or with the
polymerase (Figure 6–14C). In both series of experiments, the pattern
Chapter 6: How Cells Read the Genome: From DNA to Protein
114
(A) REACTIVE URIDINE RESIDUE
first series: U* at + 20
R
HOCH2
O
H
CH2 CH2
N
N
O
CH
H
N
RNA
chain
N
N
DNA
template
second series: U* at + 41
U*
O
N
5’ ends
N
H
NH
CH
U*
N+
C
H2C
(C) EXPERIMENTAL DATA
(B) REACTIVE TRANSCRIPTS
2
3
5
6
7
10
14
24
3
6
7
8
10
13
16
first series
second series
2 3 5 6 7 10 14 24 3 6 7 8 10 13 16
protein
DNA
3’
ends
Figure 6–14 Probing the active site of RNA polymerase (Problem 6–38). (A) The reactivity of a U analog with adenine in the
template DNA. The arrow indicates the site of attack. (B) locations of the U analog in the RNA chain. Numbers refer to the
number of nucleotides from the reactive U to the 3ʹ end. (C) The pattern of reaction of the reactive U with DNA and protein.
of reactivity with DNA and protein varied depending on the position of
the reactive analog relative to the 3ʹ end of the RNA. Why do you suppose
that the activated U reacts so strongly with DNA up to a point and then
abruptly begins to react more strongly with protein?
6–39
he intron–exon structure of eukaryotic genes came as a shock. In the
early, skeptical days, the most convincing demonstration was visual, as
shown, for example, by the electron micrograph in Figure 6–15A. his
image was obtained by hybridizing ovalbumin mRNA to a long segment
of DNA that contained the gene. To those used to looking at single- and
double-stranded nucleic acids in the electron microscope, the structure
was clear: a set of single-stranded tails and loops emanating from a central duplex segment whose ends corresponded to the ends of the mRNA
(Figure 6–15B). To the extent possible, describe the intron–exon structure of this gene.
6–40
he interaction of U1 snRNP with the sequences at the 5ʹ ends of introns
is usually shown to involve pairing between the nucleotides in the premRNA and those in the RNA component of the U1 snRNP, as illustrated
in Figure 6–16A. But given the relatively small number of bases involved,
(A) ELECTRON MICROGRAPH
(B) INTERPRETATION
5‘
end
200
nucleotides
3‘
end
Figure 6–15 Hybrid between ovalbumin mRNA and a DNA segment containing its gene (Problem 6–39). (A) An electron micrograph. (B) The
hybrid with background eliminated. Each of the loops is formed by single-stranded DNA emanating from a central RNA–DNA duplex.
FROM DNA TO RNA
115
(A) NORMAL PRE-mRNA
CAUUG
U1 snRNA
exon
5’
CAUUC
--AGGGUGAAU-- 3’
U1 snRNA
5’
splicing
CAUUU
Figure 6–16 Characterization of the pairing of U1 snRNP with pre-mRNA (Problem
6–40). (A) Interaction of U1 snRNP with a normal pre-mRNA. (B) Interaction of
U1 snRNP with a mutant pre-mRNA. (C) Mutant U1 snRNAs.
that arrangement is hardly convincing. A series of experiments tested the
hypothesis that base-pairing was critical to the function of U1 snRNP in
splicing. As shown in Figure 6–16B, a mutant pre-mRNA was generated
that could not be spliced. Several mutant U1 snRNAs were then tested
for their ability to promote splicing of the mutant pre-mRNA, with the
results shown in Figure 6–16C. Do these experiments argue that basepairing is critical to the role of U1 snRNP in splicing? If so, how?
6–41
he AAUAAA sequence just 5ʹ of the polyadenylation site is a critical signal for polyadenylation, as has been veriied in many ways. One elegant
conirmation used chemical modiication to interfere with speciic protein interactions. RNA molecules containing the signal sequence were
radiolabeled at one end and then treated with diethylpyrocarbonate,
which can modify A and G nucleotides, rendering the RNA sensitive to
breakage by subsequent aniline treatment. he experimental conditions
employed were such that the RNA molecules contained about one modiication each. hey were then cleaved with aniline. he resulting series of
fragments have lengths that correspond to the positions of As and Gs in
the RNA (Figure 6–17, lane 1).
To deine critical A and G nucleotides, the modiied, but still intact,
RNAs were incubated with an extract capable of cleavage and polyadenylation. he RNAs were then separated into those that had acquired a
poly-A tail (poly-A+) and those that had not (poly-A–). hese two fractions
were treated with aniline and the fragments were analyzed by gel electrophoresis (Figure 6–17, lanes 2 and 3). In a second reaction, EDTA was
added to the extract along with the modiied RNAs; this does not afect
RNA cleavage but prevents addition of the poly-A tail. hese cleaved
RNAs were isolated, treated with aniline, and examined by electrophoresis (lane 4).
A. At which end were the starting RNA molecules labeled?
B. Explain why the bands corresponding to the AAUAAA signal (bracket in
Figure 6–17) are missing from the poly-A+ RNA and the cleaved RNA.
C. Explain why the band at the arrow (the normal nucleotide to which poly
A is added) is missing from the poly-A+ RNA but is present in the cleaved
RNA.
D. Which A and G nucleotides are important for cleavage, and which A and
G nucleotides are important for addition of the poly-A tail?
E. What additional information might be obtained by labeling the RNA
molecules at the other end?
5’
ly
-A +
ea
ve
d
ea
3’
5’
cl
U1 snRNA
ly
CAUUA
po
(B) MUTANT PRE-mRNA
splicing
5’
po
CA UUC
--AGGGUGAGG-- 3’
tr
U1 snRNA
exon
te
d
-A –
5’
un
U1 snRNA
(C) MUTANT U1 snRNAs
WITH MUTANT PRE-mRNA
1
2
3
4
A
C
G
U
U
A
A
C
A
A
C
A
A
U
U
G
A
A
C
A
A
A
U
A
A
C
G
U
C
G
A
A
U
A
U
U
A
C
C
A
A
U
G
U
U
U
A
U
U
U
C
G
U
U
A
U
C
Figure 6–17 Autoradiographic analysis
of experiments to deine the purines
important in rna cleavage and
polyadenylation (Problem 6–41). The
sequence of the precursor rna is
shown at the left with the 5ʹ end at the
bottom and the 3ʹ end at the top. all
rnas were modiied by reaction with
diethylpyrocarbonate. rna that was
not treated with extract (untreated) is
shown in lane 1. rna that was treated
with extract but was not polyadenylated
(poly-a–) is shown in lane 2. rna that was
polyadenylated (poly-a+) in the extract is
shown in lane 3. rna that was cleaved
but not polyadenylated (cleaved) is
shown in lane 4. shorter rna fragments
run faster; that is, they travel farther
toward the bottom of the gel during
electrophoresis.
116
Chapter 6: How Cells Read the Genome: From DNA to Protein
(A) HISTONE PRECURSOR RNA
human
5’
ACCCAAAGGCUCUUUUCAGAGCCACCCAC UUAUUCCAACGAAAGUAGCUGUGAUAAUU 3’
(B) DNA OLIGONUCLEOTIDES
human
5‘
ACGAAAGTAGCTGTG 3’
mouse
5’
CGGAAAGAGCTGTT
3’
consensus
5’ AAAGAAAGAGCTGGT
3’
(C) HUMAN U7 snRNA
5’ m3G-NNGUGUUACAGCUCUUUUAGAAUUUGUCUAGU 3’
6–42
6–43
Figure 6–18 Processing of histone
precursor RNAs (Problem 6–42).
(A) Nucleotide sequences of histone
precursor RNAs. Horizontal arrows indicate
the inverted repeat sequences capable
of forming a stem-loop structure in the
precursor, the vertical arrow indicates
the site of cleavage, and the red shading
shows the position of the conserved
region. (B) DNA oligonucleotides used in
the experiments. (C) Sequence of human
U7 snRNA. The trimethylated cap, m3G,
is characteristic of “U” RNAs. N refers to
nucleotides whose identity was unknown.
Unlike most mRNAs, histone mRNAs do not have poly-A tails. hey are
processed from a longer precursor by cleavage just 3ʹ of a stem-loop
structure in a reaction that depends on a conserved sequence near
the cleavage site (Figure 6–18A). A classic paper deined the role of U7
snRNP in cleavage of the histone precursor RNA.
When nuclear extracts from human cells were treated with a nuclease
to digest RNA, they lost the ability to cleave the histone precursor. Adding back a crude preparation of snRNPs restored the activity. he extract
could also be inactivated by treatment with RNAse H (which cleaves RNA
in an RNA–DNA hybrid) in the presence of DNA oligonucleotides containing the conserved sequence—the suspected site of snRNP interaction in the histone precursor (Figure 6–18B). Somewhat surprisingly, the
mouse and mammalian consensus oligonucleotides completely blocked
histone processing, but the human oligonucleotide had no efect. he two
inhibitory oligonucleotides also caused the disappearance of a 63-nucleotide snRNA, which was then partially sequenced and identiied as U7
snRNA (Figure 6–18C).
A. Explain the design of the oligonucleotide experiment. What were these
scientists trying to accomplish by incubating the extract with a DNA oligonucleotide in the presence of RNAse H?
B. Since a human extract was used, do you ind it surprising that the human
oligonucleotide did not inhibit processing, whereas the mouse and consensus oligonucleotides did? Can you ofer an explanation for this result?
In eukaryotes, two distinct classes of snoRNAs, which are characterized by conserved sequence motifs termed boxes, are responsible for
2ʹ-O-methylation and pseudouridylation. Box C/D snoRNAs direct
2ʹ-O-ribose methylation, and box H/ACA snoRNAs direct pseudouridylation of target RNAs. Recently a novel snoRNA, called U85, was shown
to contain both box elements (Figure 6–19). You want to know how these
box elements participate in the function of U85 snoRNA. Examination
of potential substrate RNAs revealed that U85 snoRNA could potentially pair with a region of the U5 spliceosomal snRNA that carries two
2ʹ-O-methylated nucleotides, U41 and C45, and two pseudouridines,
ψ43 and ψ46 (Figure 6–20A).
To determine whether these U5 snRNA sequences were true substrates for modiication by U85 snoRNA, the relevant segment of U5 was
inserted into a region of the U2 snRNA gene to create a distinctive U2–
U5 hybrid molecule that could be readily followed. In addition, a second hybrid molecule, U2–U5m, was constructed to contain a mutant
segment of U5, and a mutant U85 snoRNA, U85m, was generated with
compensating changes in the guide sequences adjacent to the box H
and box D regions, so that it could pair with U2–U5m (Figure 6–20B).
When expression vectors for U2–U5, U2–U5m, and U85m were transfected into cells, the encoded RNAs were shown to accumulate normally.
2ʹ-O-methylation and pseudouridylation in the critical region in the
H
ACA
C’
D’
D’
C
D
Figure 6–19 Proposed secondary structure
of the human U85 snoRNA (Problem 6–43).
Box elements are indicated by boxes.
Regions that are thought to be important
for targeting 2ʹ-O-methylation and
pseudouridylation are highlighted in yellow.
FROM DNA TO RNA
117
TABLE 6–2 Modification of potential target nucleotides in various transfections
(Problem 6–43).
Modiicationsb
pseudouridine
2ʹ-O-Methyl
U43
U46
U41
C45
–
ψ
m
m
+
–
–
m
–
+
–
ψ
m
m
Transfected molecules
U85a
1
+
2
+
3
+
U85m
U2–U5
U2–U5m
+
+
aEndogenous U85 is present in all cells.
bNucleotides that have been converted to pseudouridine are indicated by ψ;
nucleotides that have been 2ʹ-O-methylated are indicated by an m.
U2–U5 and U2–U5m molecules were detected by sequencing, as summarized in Table 6–2.
What are the expectations of these experiments if bases in U5 snRNA
serve as bona ide targets for U85 snoRNA-dependent modiication?
Which, if any, of the naturally modiied nucleotides in U5—methylated riboses at U41 and C45, and the pseudouridines ψ43 and ψ46—are
dependent on U85 snoRNA? Explain your reasoning.
MEDICAL LINKS
6–44
he trypanosome, which is the microorganism that causes sleeping sickness, can vary its surface glycoprotein coat and thus evade the immune
defenses of its host. he promoter for the variable surface glycoprotein
(VSG) gene proved diicult to locate, but was mapped by measuring
the sensitivity of the transcript to ultraviolet (UV) irradiation. Since RNA
polymerases cannot transcribe through pyrimidine dimers (the major
damage produced by UV irradiation), the sensitivity of transcription to
UV irradiation is a measure of the distance between the start of transcription (the promoter) and the point at which transcription is assayed (Vsg
gene).
Transcription through rRNA genes was used to calibrate the system.
he 5S RNA transcription unit is just over 100 nucleotides long, whereas
the 18S, 5.8S, and 28S rRNAs are part of a single transcription unit that
is about 8 kb in length (Figure 6–21A). Trypanosomes were exposed to
increasing doses of UV irradiation; their nuclei were then isolated and
incubated with 32P-NTPs to radioactively label the RNA. RNA isolated
from the nuclei was hybridized to DNA probes corresponding to the 5S
RNA gene and various parts of the rRNA gene (Figure 6–21B). Plots of
the logarithm of the counts in each spot against the UV dose gave straight
lines (Figure 6–21C), with slopes that were proportional to the distance
from the hybridization probe to the promoter.
When the experiment was done with a probe from the beginning of
the Vsg gene, transcription was found to be inactivated about seven times
faster for the Vsg gene than for probe 4 from the ribosomal transcription
unit.
A. Why does RNA transcription increase in sensitivity to UV irradiation with
increasing distance from the promoter?
B. Roughly how far is the Vsg gene from its promoter? What assumption do
you have to make in order to estimate this distance?
C. Transcription through another gene, located about 10 kb upstream of the
Vsg gene, was about 20% less sensitive to UV irradiation than transcription
(A) WILD TYPE
U85
G C
G U
AAUCUU GUAAAGGGG box H
U2–U5
UUUAGAAAψCAψUUUC –39
U85
AGAUCUUUGGUAAU box D
(B) MUTANT
U85m
G C
G U
AUAGAA GUAAAGGGG box H
U2–U5m UUAUCUUAψCAψUUUC –39
U85m
AGUAGAAUGGUAAU box D
Figure 6–20 analysis of potential rna
targets for modiication by U85 snorna
(Problem 6–43). (a) potential pairing
between the guide sequences in U85
snorna and a segment of U5 in U2–U5
snrna. pairing with the guide sequence
adjacent to box h, which typically directs
pseudouridylation, is shown above
U2–U5. pairing with the guide sequence
adjacent to box D, which typically directs
2ʹ-O-methylation, is shown below U2–U5.
Only one pairing could occur at a time.
pseudouridines are indicated with ψ, and
sites of 2ʹ-O-methylation are indicated
with dots. (B) potential pairing between the
mutant guide sequences in U85m snorna
and the mutant segment of U5 in U2–U5m.
118
Chapter 6: How Cells Read the Genome: From DNA to Protein
Figure 6–21 UV mapping of the promoter
in trypanosomes (Problem 6–44).
(A) Structure of the ribosomal RNA
transcription unit. The positions of the
hybridization probes along with a scale
marker are shown. The transcription
unit begins at the left end of the arrow.
(B) A dot blot of transcripts from the
5S RNA and rRNA genes. The dot blot
is done by placing an excess of the DNA
probe in spots on a ilter paper and then
hybridizing radiolabeled rna to it.
(C) sensitivities of the transcription units
to increasing doses of UV irradiation.
(A) TRANSCRIPTION MAP
18S
1
5.8S
28S
2
3
4
1 kb
probes
(C) SENSITIVITY TO UV
transcription (% of control)
(B) UV DOSE RESPONSE
0 UV dose
5S RNA
rRNA 1
rRNA 2
rRNA 3
rRNA 4
5S gene
100
1
2
10
3
4
1
0
200
400
600
800
UV dose (ergs/mm2)
through the Vsg gene. Is this measurement consistent with the possibility
that these two genes are transcribed from the same promoter? Explain
your reasoning.
FROM RNA TO PROTEIN
TERMS TO LEARN
aminoacyl-tRNA
synthetase
anticodon
codon
eukaryotic initiation
factor (eIF)
genetic code
induced fit
initiator tRNA
kinetic proofreading
molecular chaperone
nonsense-mediated
mRNA decay
proteasome
reading frame
ribosome
ribozyme
rRNA (ribosomal
RNA)
translation
tRNA (transfer RNA)
DEFINITIONS
Match the deinition below with its term from the list above.
6–45
Large protein complex in the cytosol and nucleus with proteolytic activity that is responsible for degrading the proteins marked for destruction.
6–46
Set of rules specifying the correspondence between nucleotide triplets in
DNA or RNA and amino acids in proteins.
6–47
Special tRNA that carries methionine and is used to begin translation.
6–48
Sequence of three nucleotides in a tRNA that is complementary to a
three-nucleotide sequence in an mRNA molecule.
6–49
RNA molecule with catalytic activity.
6–50
Surveillance system in eukaryotes that eliminates defective mRNAs
before they can be translated into protein.
6–51
he three-nucleotide phase in which nucleotides in an mRNA are translated into amino acids in a protein.
FROM RNA TO PROTEIN
6–52
Enzyme that attaches the correct amino acid to a tRNA molecule to form
the activated intermediate used in protein synthesis.
6–53
Protein that helps other proteins fold correctly.
119
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
6–54
Wobble pairing occurs between the irst position in the codon and the
third position in the anticodon.
6–55
During protein synthesis, the thermodynamics of base-pairing between
tRNAs and mRNAs sets the upper limit for the accuracy with which protein molecules are made.
THOUGHT PROBLEMS
6–56
For the RNA sequence below, indicate the amino acids that are encoded
in the three reading frames. If you were told that this segment of RNA was
in the middle of an mRNA that encoded a large protein, would you know
which reading frame was used? How so? (he genetic code is shown in
Table 7 on page 966.)
AGUCUAGGCACUGA
6–57
After treating cells with a chemical mutagen, you isolate two mutants.
One carries alanine and the other carries methionine at a site in the protein that normally contains valine (Figure 6–22). After treating these two
mutants again with the mutagen, you isolate mutants from each that now
carry threonine at the site of the original valine (Figure 6–22). Assuming that all mutations involve single-nucleotide changes, deduce the
codons that are used for valine, methionine, threonine, and alanine at
the afected site. Would you expect to be able to isolate a valine-to-threonine mutant in one step?
6–58
he genetic code was deciphered in part by experiments in which polynucleotides of repeating sequences were used as mRNAs to direct protein synthesis in cell-free extracts. In the test tube, artiicial conditions
were used that allowed ribosomes to start protein synthesis anywhere
on an RNA molecule, without the need for a translation start codon, as
required in a living cell. What polypeptides would you expect to be synthesized if the following polynucleotides were used as templates in such
a cell-free extract?
A. UUUUUUUUUUUU…
B. AUAUAUAUAUAU…
C. AUCAUCAUCAUC…
6–59
In B. licheniformis, a few amino acids are removed from the C-terminus of
the β-lactamase enzyme after it is synthesized. he sequence of the original C-terminus can be deduced by comparing it to a mutant in which the
reading frame is shifted by the insertion or deletion of a nucleotide and
the mutant β-lactamase escapes cleavage. he amino acid sequences of
the puriied wild-type enzyme and the frameshift mutant from amino
acid 263 to the C-terminal end are given below.
↓ site of protein cleavage
wild type:
N M N G K
frameshift mutant: N M I W Q I C V M K D
A. What was the mutational event that gave rise to the frameshift mutant?
B. Deduce the number of amino acids in the synthesized form of the
wild-type enzyme and, as far as possible, the sequence of the deleted
C-terminus.
first
treatment
Ala
Val
second
treatment
Thr
Met
Figure 6–22 Two rounds of mutagenesis
and the altered amino acids at a single
position in a protein (Problem 6–57).
120
Chapter 6: How Cells Read the Genome: From DNA to Protein
6–60
Which of the following mutational changes would you predict to be the
most deleterious to gene function? Explain your answers.
1. Insertion of a single nucleotide near the end of the coding sequence.
2. Removal of a single nucleotide near the beginning of the coding sequence.
3. Deletion of three consecutive nucleotides in the middle of the coding
sequence.
4. Deletion of four consecutive nucleotides in the middle of the coding
sequence.
5. Substitution of one nucleotide for another in the middle of the coding
sequence.
6–61
Consider the properties of two hypothetical genetic codes constructed
with the four common nucleotides: A, G, C, and T.
A. Imagine that one genetic code is constructed so that pairs of nucleotides
are used as codons. How many diferent amino acids could such a code
specify?
B. Imagine that the other genetic code is a triplet code; that is, it uses three
nucleotides to specify each amino acid. In this code, the amino acid speciied by each codon depends only on the composition of the codon—not
the sequence. hus, for example, CCA, CAC, and ACC, which all have the
composition C2A, would encode the same amino acid. How many diferent amino acids could such a code specify?
C. Would you expect the genetic codes in A and B to lead to diiculties in
the process of translation, using mechanisms analogous to those used in
translating the standard genetic code?
6–62
One remarkable feature of the genetic code is that amino acids with similar chemical properties often have similar codons. Codons with U or C as
the second nucleotide, for example, tend to specify hydrophobic amino
acids. Can you suggest a possible explanation for this phenomenon in
terms of the early evolution of the protein-synthesis machinery?
6–63
he rules for wobble base-pairing in bacteria and eukaryotes are shown
in Table 6–3. On the left side of the table, the rules are expressed as a
wobble codon base and its recognition by possible anticodon bases. [he
anticodon base I (inosine) is a common modiication in tRNAs; it is generated by deamination of A.] Reformulate these rules as particular anticodon bases and their recognition by possible codon bases, as suggested
by the partial information on the right side of the table.
TABLE 6–3 Rules for wobble base-pairing between codon and anticodon
(Problem 6–63).
Wobble
codon
base
Possible
anticodon
base
Bacteria
U
C
A
G
A, G, or I
G or I
U or I
C or U
Bacteria
U
C
A
G
I
Eukaryotes
U
C
A
G
G or I
G or I
U
C
Eukaryotes
U
C
A
G
I
Wobble
anticodon
base
Possible
codon
base
FROM RNA TO PROTEIN
6–64
Given the wobble rules for codon–anticodon pairing in bacteria, the minimum number of diferent tRNAs that would be required to recognize all
61 codons is 31. What is the minimum number of diferent tRNAs that is
consistent with the wobble rules used in eukaryotes (see Table 6–3)?
6–65
A mutation in a bacterial gene generates a UGA stop codon in the middle
of the mRNA coding for the protein product. A second mutation in the
cell leads to a single nucleotide change in a tRNA that allows the correct
translation of the protein; that is, the second mutation “suppresses” the
defect caused by the irst. he altered tRNA translates the UGA codon
as tryptophan. What nucleotide change has probably occurred in the
mutant tRNA molecule? What consequences would the presence of such
a mutant tRNA have for the translation of the normal genes in this cell?
6–66
In a clever experiment performed in 1962, a cysteine that was already
attached to tRNACys was chemically converted to an alanine. hese alanyl-tRNACys molecules were then added to a cell-free translation system
from which the normal cysteinyl-tRNACys molecules had been removed.
When the resulting protein was analyzed, it was found that alanine had
been inserted at every point in the protein chain where cysteine was supposed to be. Discuss what this experiment tells you about the role of aminoacyl-tRNA synthetases during the normal translation of the genetic
code.
6–67
he charging of a tRNA with an amino acid occurs according to the reaction
amino acid + tRNA + ATP → aminoacyl-tRNA + AMP + PPi
where PPi is pyrophosphate, the linked phosphates that were cleaved
from ATP to generate AMP. In the aminoacyl-tRNA, the amino acid and
tRNA are linked by a high-energy bond. hus, a large portion of the energy
derived from the hydrolysis of ATP is stored in this bond and is available
to drive peptide bond formation at the later stages of protein synthesis.
he free-energy change (ΔG°) for the charging reaction shown above is
close to zero, so that attachment of the amino acid to tRNA would not be
expected to be dramatically favored. Can you suggest a further step that
could help drive the charging reaction to completion?
6–68
he protein you are studying contains ive leucines and consists of a single polypeptide chain. One leucine is C-terminal and another is N-terminal. In a suspension of cells, the average time required to synthesize this
polypeptide is 8 minutes. At time zero, radioactive leucine is added to ive
diferent suspensions of cells that are already in the process of synthesizing the protein. You isolate the complete protein from individual suspensions at 2, 4, 6, 8, and 80 minutes. (Any incomplete polypeptide chains
are eliminated at this step.) he proteins are then analyzed for N-terminal and total radioactive leucine. With increasing time of exposure of the
cells to the radioactive leucine, the ratio of N-terminal radioactivity to
total radioactivity in the isolated protein should:
A. Increase to a inal value of 0.2.
B. Remain constant at a value of 0.2.
C. Decrease to a inal value of 0.2.
D. An answer cannot be determined from this information.
6–69
It is commonly reported that 30–50% of a cell’s energy budget is spent on
protein synthesis. How do you suppose such a measurement might be
made?
6–70
One strand of a section of DNA isolated from E. coli reads
5ʹ-GTAGCCTACCCATAGG-3ʹ
121
Chapter 6: How Cells Read the Genome: From DNA to Protein
A. Suppose that an mRNA were transcribed using the complement of this
DNA strand as the template. What would the sequence of the mRNA in
this region be?
B. How many diferent peptides could potentially be made from this
sequence of RNA? Would the same peptides be made if the other strand
of the DNA served as the template for transcription?
C. What peptide would be made if translation started exactly at the 5ʹ end of
the mRNA in part A? When tRNAAla leaves the ribosome, what tRNA will
be bound next? When the amino group of alanine forms a peptide bond,
which bonds, if any, are broken, and what happens to tRNAAla?
6–71
Polycistronic mRNAs are common in prokaryotes but extremely rare in
eukaryotes. Describe the key diferences in protein synthesis that underlie this observation.
6–72
Prokaryotes and eukaryotes both protect against the dangers of translating broken mRNAs. What dangers do partial mRNAs pose for the cell?
6–73
he antibiotic edeine inhibits protein synthesis but has no efect on
either DNA synthesis or RNA synthesis. When added to a reticulocyte lysate, edeine stops protein synthesis after a short lag, as shown in
Figure 6–23. By contrast, cycloheximide stops protein synthesis immediately (Figure 6–23). Analysis of the edeine-inhibited lysate by densitygradient centrifugation showed that no polyribosomes remained at the
time protein synthesis had stopped. Instead, all the globin mRNA accumulated in an abnormal 40S peak, which contained equimolar amounts
of the small ribosomal subunit and initiator tRNA.
A. What step in protein synthesis does edeine inhibit?
B. Why is there a lag between addition of edeine and cessation of protein
synthesis? What determines the length of the lag?
C. Would you expect the polyribosomes to disappear if you added cycloheximide at the same time as edeine?
6–74
In a reticulocyte lysate, the polynucleotide 5ʹ-AUGUUUUUUUUU directs
the synthesis of Met–Phe–Phe–Phe. In the presence of farsomycin, a new
antibiotic perfected by Fluhardy Pharmaceuticals, this polymer directs
synthesis of Met–Phe only. From this information, which of the following
deductions could you make about farsomycin?
A. It prevents formation of the 80S initiation complex, which contains the
initiator tRNA and both ribosomal subunits.
B. It inhibits binding of aminoacyl-tRNAs to the A site in the ribosome.
C. It inactivates peptidyl transferase activity of the large ribosomal subunit.
D. It blocks translocation of peptidyl-tRNA from the A site to the P site of the
ribosome.
E. It interferes with chain termination and release of the peptide.
6–75
Both hsp60-like and hsp70 molecular chaperones share an ainity for
exposed hydrophobic patches on proteins, using them as indicators of
incomplete folding. Why do you suppose hydrophobic patches serve as
critical signals for the folding status of a protein?
6–76
Most proteins require molecular chaperones to assist in their correct
folding. How do you suppose the chaperones themselves manage to fold
correctly?
6–77
Your advisor, the brilliant bioinformatician, has a high regard for your
intellect and industry. She suggests that you write a computer program
that will identify the exons of protein-encoding genes directly from
the sequence of the human genome. In preparation for that task, you
decide to write down a list of the features that might distinguish coding
sequences from intronic DNA and sequences outside of genes. What features would you list?
radioactivity in hemoglobin
122
control
add
inhibitor
edeine
cycloheximide
0
2
4
6
time (minutes)
8
Figure 6–23 Effects of the inhibitors
edeine and cycloheximide on protein
synthesis in reticulocyte lysates
(Problem 6–73).
FROM RNA TO PROTEIN
123
5
10
5
10
15
20
25
30
25
30
kd
116
97
68
56
Figure 6–24 Time course of synthesis
of a TMV protein in a rabbit-reticulocyte
lysate (Problem 6–78). No radioactivity
was detected during the irst 3 minutes
because the short chains ran off the
bottom of the gel. sDs denatures
proteins so that they run approximately
according to their molecular masses.
a scale of molecular masses in
kilodaltons is shown on the left.
40
31
20
15
20
time of sample (minutes)
CALCULATIONS
6–79
he average molecular weight of proteins encoded in the human genome
is about 50,000. A few proteins are very much larger than this average.
For example, the protein called titin, which is made by muscle cells, has
a molecular weight of 3,000,000.
A. Estimate how long it will take a muscle cell to translate an mRNA coding
for an average protein and one coding for titin. he average molecular
mass of amino acids is about 110 daltons. Assume that the translation
rate is two amino acids per second.
B. If the nucleotides in the coding portion of the mRNA constitute 5% of the
total that are transcribed, how long will it take a muscle cell to transcribe
a gene for an average protein versus the titin gene. Assume that the transcription rate is 20 nucleotides per second.
(A)
molecular mass (kd)
Rates of peptide-chain growth can be estimated from data such as those
shown in Figure 6–24. In this experiment, a tobacco mosaic virus (TMV)
mRNA, which encodes a 116,000 dalton protein, was translated in a
rabbit-reticulocyte lysate in the presence of 35S-methionine. Samples
were removed at one-minute intervals and subjected to electrophoresis
on SDS-polyacrylamide gels. he separated translation products were
visualized by autoradiography. As is apparent in Figure 6–24, the largest
detectable polypeptides get larger with time, until the full-length protein
appears at about 25 minutes.
A. Is the rate of synthesis linear with time? One simple way to answer this
question is to determine the molecular mass of the largest peptide in
each sample, as determined by reference to the standards shown on the
left in Figure 6–24, and then plot each of these masses against the time at
which the relevant sample was taken.
B. What is the rate of protein synthesis (in amino acids/minute) in this
experiment? Assume the average molecular mass of an amino acid is 110
daltons.
C. Why does the autoradiograph have so many bands in it rather than just a
few bands that get larger as time passes; that is, why does the experiment
produce the “actual” result (Figure 6–25A) rather than the “theoretical”
result (Figure 6–25B)? Can you think of a way to manipulate the experimental conditions to produce the theoretical result?
ACTUAL RESULT
116
68
56
40
30
time
(B)
molecular mass (kd)
6–78
THEORETICAL RESULT
116
68
56
40
30
time
Figure 6–25 potential outcomes of
experiments on rates of protein synthesis
(Problem 6–78).
124
Chapter 6: How Cells Read the Genome: From DNA to Protein
6–80
Protein synthesis consumes four high-energy phosphate bonds per
added amino acid. Transcription consumes two high-energy phosphate
bonds per added nucleotide. Calculate how many protein molecules will
have been made from an individual mRNA at the point when the energy
cost of translation is equal to the energy cost of transcription. Assume
that the nucleotides in the coding portion of the mRNA constitute 5% of
the total that are transcribed.
6–81
he overall accuracy of protein synthesis is diicult to measure because
mistakes are very rare. One ingenious approach used lagellin (molecular weight 40,000). Flagellin is the sole protein in bacterial lagella and
thus easy to purify. Because lagellin contains no cysteine, it allows for
sensitive detection of cysteine that has been misincorporated into the
protein.
To generate radioactive cysteine, bacteria were grown in the presence
of 35SO42– (speciic activity 5.0 × 103 cpm/pmol) for exactly one generation with excess unlabeled methionine in the growth medium (to prevent the incorporation of 35S label into methionine). Flagellin was puriied and assayed: 8 μg of lagellin was found to contain 300 cpm of 35S
radioactivity.
A. Of the lagellin molecules that were synthesized during the labeling
period, what fraction contained cysteine? Assume that the mass of lagellin doubles during the labeling period and that the speciic activity of
cysteine in lagellin is equal to the speciic activity of the 35SO42– used to
label the cells.
B. In lagellin, cysteine is misincorporated at the arginine codons CGU and
CGC. In terms of anticodon–codon interaction, what mistake is made
during the misincorporation of cysteine for arginine?
C. Given that there are 18 arginines in lagellin, and assuming that all
arginine codons are equally represented, what is the frequency of misreading of each sensitive (CGU and CGC) arginine codon?
D. Assuming that the error frequency per codon calculated above applies to
all amino acid codons equally, calculate the percentage of molecules that
are correctly synthesized for proteins 100, 1000, and 10,000 amino acids
in length. he probability of synthesizing a correct protein is P = (1 – E)n,
where E is the error frequency and n is the number of amino acids added.
DATA HANDLING
6–82
Many of the errors in protein synthesis occur because tRNA synthetases
have diiculty discriminating between related amino acids. For example,
isoleucyl-tRNA synthetase (IleRS) normally activates isoleucine.
IleRS + Ile + ATP → IleRS(Ile-AMP) + PPi
At a frequency about 1/180 of the correct activation, IleRS misactivates
valine.
IleRS + Val + ATP → IleRS(Val-AMP) + PPi
Protein synthesis is more accurate than this frequency might suggest
because the synthetase subsequently edits out most of its mistakes, in a
reaction that depends on the presence of tRNAIle.
IleRS(Val-AMP) + tRNAIle → IleRS + Val + AMP + tRNAIle (EDITING)
he tRNAIle is, of course, also required for proper aminoacylation (charging) by isoleucine to make Ile-tRNAIle.
IleRS(Ile-AMP) + tRNAIle → Ile-tRNAIle + IleRS + AMP
(CHARGING)
Are the parts of tRNAIle that are required for isoleucine charging the same
as those that are required for valine editing?
FROM RNA TO PROTEIN
125
(A) tRNA STRUCTURES
(B) tRNA CHARGING AND EDITING
Val
tRNA
(percent)
tRNA
Ile
activity relative to tRNA
TΨC
loop
D loop
UAC
anticodon
loop
100
isoleucine charging
valine editing
Ile
acceptor
stem
50
0
GAU
UAC
GAU
GAU
GAU
GAU
GAU
GAU
GAU
Figure 6–26 Portions of tRNAIle responsible for charging and editing (Problem 6–82). (A) tRNAVal and tRNAIle. (B) Isoleucine
charging and valine editing. Chimeric tRNAs composed of bits from tRNAVal (green) and tRNAIle (blue) are shown below the
results for isoleucine charging (red bars) and valine editing (yellow bars).
One approach to this question is to make changes in the tRNAIle to
see whether the two activities track with one another. Rather than change
the sequence nucleotide by nucleotide, blocks of sequence changes
were made, using tRNAVal as a donor. tRNAVal by itself does not stimulate
isoleucine charging or valine editing by IleRS. Changing its anticodon
from 5ʹ-CAU to 5ʹ-GAU, however, allows it to be charged fairly eiciently
by IleRS. A variety of chimeric tRNAs were made by combining bits of
tRNAIle and tRNAVal. he ability of each chimera to stimulate isoleucine
charging and valine editing was then tested, as shown in Figure 6–26.
A. What (at a minimum) must be inserted into tRNAVal to permit isoleucine
charging by IleRS?
B. What (at a minimum) must be inserted into tRNAVal to permit valine editing by IleRS?
C. Does IleRS recognize the same features of tRNAIle when it catalyzes isoleucine charging that it does when it carries out valine editing? Explain
your answer.
6–83
Consider the following experiment on the coordinated synthesis of
the α and β chains of hemoglobin. Rabbit reticulocytes were labeled
with 3H-lysine for 10 minutes, which is very long relative to the time
required for the synthesis of a single globin chain. he ribosomes, with
attached nascent globin chains, were then isolated by centrifugation to
give a preparation free of soluble (inished) globin chains. he nascent
globin chains were digested with trypsin, which gives peptides ending in
C-terminal lysine or arginine. he peptides were then separated by highperformance liquid chromatography (HPLC), and their radioactivity was
measured. A plot of the radioactivity in each peptide versus the position
of the lysines in the chains (numbered from the N-terminus) is shown in
Figure 6–27.
A. Do these data allow you to decide which end of the globin chain (N- or
C-terminus) is synthesized irst? How so?
B. In what ratio are the two globin chains produced? Can you estimate the
relative numbers of α- and β-globin mRNA molecules from these data?
C. How long does a protein chain stay attached to the ribosome once the
termination codon has been reached?
D. It was once suggested that heme is added to nascent globin chains during their synthesis and, furthermore, that ribosomes must wait for the
126
Chapter 6: How Cells Read the Genome: From DNA to Protein
radioactivity
in peptide
β chain
1000
α chain
0
20
40
60
80
100
120
radioactivity
in peptide
Figure 6–27 Synthesis of α- and β-globin chains
(Problem 6–83).
insertion of heme before they can proceed. he straight lines in Figure
6–27 indicate that ribosomes do not pause signiicantly, and heme is
now thought to be added after synthesis. From among the graphs shown
in Figure 6–28, choose the one that would have resulted if there were a
signiicant roadblock to ribosome movement halfway down the globin
mRNA.
6–84
6–85
Termination codons in bacteria are decoded by one of two proteins.
Release factor 1 (RF1) recognizes UAG and UAA, whereas RF2 recognizes
UGA and UAA. For RF2, a comparison of the nucleotide sequence of the
gene with the amino acid sequence of the protein revealed a startling
surprise, which is contained within the sequences shown below the gene
in Figure 6–29. Sequences of the gene and protein were checked carefully to rule out any artifacts.
A. What is the surprise?
B. What hypothesis concerning the regulation of RF2 expression is suggested by this observation?
You are studying protein synthesis in Tetrahymena, which is a unicellular ciliate. You have good news and bad news. he good news is that
you have the irst bit of protein and nucleic acid sequence data for the
C-terminus of a Tetrahymena protein, as shown below:
I
M
Y
K
Q
V
A
Q
T
Q
L
*
AUU AUG UAU AAG UAG GUC GCA UAA ACA CAA UUA UGA GAC UUA
he bad news is that you have been unable to translate a preparation
of Tetrahymena mRNA in a reticulocyte lysate, which is a standard system for analyzing protein synthesis in vitro. he mRNA preparation
looks good by all criteria, but the translation products are mostly small
polypeptides (Figure 6–30, lane 1).
A P G *
GCACCGGGGTAG
M F E I
ATGTTTGAAATT
1
500
1000
(B)
140
residue number
1500
GGGTATCTTTGACTACGACGCC
G Y L D Y D A
Figure 6–29 Schematic representation of the gene for RF2 (Problem 6–84). The
coding sequence is shown as a blue line, with sequences at the start and inish
shown for reference.
(C)
radioactivity
in peptide
radioactivity in peptide (cpm)
(A)
2000
residue number
Figure 6–28 Hypothetical curves for
globin synthesis with a roadblock to
ribosome movement at the midpoint
of the mRNA (Problem 6–83). These
schematic diagrams are analogous to
the graph in Figure 6–27.
FROM RNA TO PROTEIN
127
Figure 6–30 Translation of TMV and Tetrahymena mRNA in a reticulocyte
lysate in the presence and absence of various components from
Tetrahymena (Problem 6–85). The molecular masses of marker proteins
are indicated in kilodaltons on the left.
Tetrahymena
RNA
Tetrahymena
cytoplasm
TMV mRNA
+
–
+
+
+
–
–
–
+
+
–
+
+
+
–
1
2
3
4
5
kd
To igure out what is wrong, you do a number of control experiments
using a pure mRNA from tobacco mosaic virus (TMV) that encodes a
116 kd protein. TMV mRNA alone is translated just ine in the in vitro system, giving a major band at 116 kd—the expected product—and a very
minor band about 50 kd larger (Figure 6–30, lane 2). When Tetrahymena
RNA is added, there is a decrease in the smaller of the two bands and a
signiicant increase in the larger one (lane 3). When some Tetrahymena
cytoplasm (minus the ribosomes) is added, the TMV mRNA now gives
mostly the higher molecular mass product (lane 4); furthermore, much
to your delight, the previously inactive Tetrahymena mRNA now appears
to be translated (lane 4). You conirm this by leaving out the TMV mRNA
(lane 5).
A. What is unusual about the sequence data for the Tetrahymena protein?
B. How do you think the more minor of the two higher molecular mass
bands is produced from pure TMV mRNA in the reticulocyte lysate (lane
2)?
C. Explain the basis for the shift in proportions of the major and minor TMV
proteins upon addition of Tetrahymena RNA alone and in combination
with Tetrahymena cytoplasm. What Tetrahymena components are likely
to be required for the eicient translation of Tetrahymena mRNA?
D. Comment on the evolutionary implications of your results.
6–86
Hsp70 molecular chaperones are thought to bind to hydrophobic regions
of nascent polypeptides on ribosomes. his binding was diicult to demonstrate for dnaK, which is one of the two major hsp70 chaperones in
E. coli. In one approach, nascent proteins were labeled with a 15-second pulse of 35S-methionine, isolated in the absence of ATP, and then
incubated with antibodies against dnaK. A collection of labeled proteins
was precipitated as shown in Figure 6–31A, lane 1. he proteins were not
precipitated if they were treated beforehand with the strong detergent
SDS (lane 2), or if they were isolated from a mutant strain missing dnaK
(dnaK-deletion, lane 3). If labeled dnaK-deletion cells were mixed with
unlabeled wild-type cells before the proteins were isolated, dnaK antibodies did not precipitate labeled proteins (lane 4). Finally, if unlabeled
methionine was added in excess after the pulse of 35S-methionine, the
labeled proteins disappeared with time (Figure 6–31B).
A. Do the series of control experiments in Figure 6–31A, lanes 2 to 4, argue
that dnaK is bound to the labeled proteins in a meaningful way (as
opposed to a random aggregation, for example)?
kd (A) PULSE
97
66
45
36
29
24
(B) CHASE
dnaK
20
14
1 2 3 4
lane number
0.25 0.5 1 2 4 6 8 10
time (minutes)
220
116
94
68
40
31
22
Figure 6–31 Association of dnak
with nascent proteins (Problem
6–86). (A) Pulse-labeled proteins
immunoprecipitated by antibodies
to dnak. Wild-type bacteria (lane 1);
SDS-treated wild-type bacteria (lane
2); dnaK-deletion strain (lane 3); and
mixture of labeled dnaK-deletion strain
and unlabeled wild-type strain (lane 4).
Size markers are indicated on the left
and the position of dnak is indicated on
the right. (B) Pulse-chase experiment.
Wild-type bacteria labeled for 15 seconds
with 35S-methionine were then incubated
for varying times in the presence of an
excess of unlabeled methionine before
immunoprecipitation by antibodies
against dnak.
128
Chapter 6: How Cells Read the Genome: From DNA to Protein
incorrectly
folded
protein
hydrophobic
protein-binding
sites
correctly
folded
protein
GroES cap
ATP
ADP
+ Pi
hsp60-like
protein complex
Figure 6–32 Protein refolding by the
bacterial GroEl chaperone (Problem
6–87). A misfolded protein is initially
captured by hydrophobic interactions
along one rim of the barrel. The
subsequent binding of ATP plus the
GroES cap increases the size of the
cavity and conines the protein in the
enclosed space, where it has a new
opportunity to fold. after about 15
seconds, aTp hydrolysis ejects the
protein, whether folded or not, and
the cycle repeats.
B. When ATP was present during the isolation of the proteins, antibodies
against dnaK did not precipitate any proteins. How do you suppose ATP
might interfere with precipitation of labeled proteins?
C. Why do you suppose that the labeled proteins disappeared with time in
the presence of excess unlabeled methionine?
D. Do these experiments show that dnaK binds to proteins as they are being
synthesized on ribosomes? Why or why not?
Hsp60-like molecular chaperones provide a large central cavity in which
misfolded proteins can attempt to refold. Two models, which are not
mutually exclusive, can be considered for the role of the hsp60-like chaperones in the refolding process. hey might act passively to provide an
isolation chamber that aids protein folding by preventing aggregation
with other proteins. Alternatively, hsp60-like chaperones might actively
unfold misfolded proteins to remove stable, but incorrect, intermediate
structures that block proper folding. he involvement of ATP binding and
hydrolysis and the associated conformational changes of the hsp60-like
chaperones could be used in favor of either model.
In bacteria, the hsp60-like chaperone GroEL binds to a misfolded
protein, then binds ATP and the GroES cap, and after about 15 seconds
hydrolyzes the ATP and ejects the protein (Figure 6–32). To distinguish
between a passive and an active role for GroEL in refolding, you label a
protein by denaturing it in tritiated water, 3H2O. When the denaturant
is removed and the protein is transferred to normal water, 1H2O, most
of the radioactivity is lost within 10 minutes, but a stable core of 12 tritium atoms exchanges on a much longer time scale—a behavior typical
of amide hydrogen atoms involved in stable hydrogen bonds. Disruption
of these bonds would allow their exchange within a few milliseconds.
You prepare the radioactive substrate and mix it immediately with a
slight molar excess of GroEL, and then wait 10 minutes for the rapidly
exchanging tritium atoms to be lost. he addition of GroES or ATP alone
has no efect on the exchange, as shown by the upper curve in Figure
6–33. he addition of GroES and ATP together causes a rapid loss of
tritium (Figure 6–33, lower curve). Addition of GroES plus ADP has no
30
tritium remaining
(atoms/picomole)
6–87
20
addition
GroES or ATP
10
GroES + ATP
0
10
20
30
40
time (minutes)
50
60
Figure 6–33 effects of chaperone
components on rates of tritium exchange
(Problem 6–87).
FROM RNA TO PROTEIN
129
Figure 6–34 Fusion proteins encoded
by three yeast plasmids (Problem 6–88).
Upon expression in yeast, the fusion
K proteins are cleaved at the peptide
K bonds indicated by the arrow. The
β-galactosidase molecules liberated by
K cleavage differ only at their N-termini.
cleavage
β-galactosidase
ubiquitin
M
G M
M
G I
M
G R
78 residues
1045 residues
efect, but GroES plus AMPPNP, a nonhydrolyzable analog of ATP, promotes a rapid exchange that is indistinguishable from GroES plus ATP.
A. After the addition of components to the complex of tritiated protein and
GroEL, it took a minimum of 45 seconds to separate the protein from the
freed tritium label. Did the exchange of tritium occur within one cycle of
binding and ejection by GroEL, which takes about 15 seconds, or might it
have required more than one cycle? Explain your answer.
B. Do the results support a passive isolation-chamber model, or an activeunfolding model, for GroEL action? Explain your reasoning.
he life-spans of proteins are appropriate to their in vivo tasks: structural
proteins tend to be long-lived; regulatory proteins are usually short-lived.
In eukaryotic cells, life-spans are strongly inluenced by the N-terminal
amino acid. he irst experiments that revealed these efects used hybrid
proteins consisting of ubiquitin fused to β-galactosidase, as shown in
Figure 6–34. When plasmids encoding these proteins were introduced
into yeast, the hybrid proteins were synthesized but the ubiquitin was
cleaved of exactly at the junction with β-galactosidase, generating proteins with diferent N-termini (Figure 6–34).
To measure the half-lives of these β-galactosidase molecules, yeast
cells containing the plasmids were grown for several generations in the
presence of a radioactive amino acid. Protein synthesis was then blocked
with the inhibitor cycloheximide (CHX). he rate of degradation of
β-galactosidase was determined by removing samples from the cultures
at various times, purifying β-galactosidase, and measuring the amount of
associated radioactivity after SDS-gel electrophoresis. he results at the
5-minute time point are shown in Figure 6–35A, and a graph depicting
results for all time points is shown in Figure 6–35B.
A. Using recombinant DNA techniques, it would have been straightforward to generate a series of plasmids in which the irst codon in the
β-galactosidase gene was changed. Why do you suppose this more direct
approach was not tried?
B. Estimate the half-life (time at which half the material has been degraded)
of each of the three species of β-galactosidase.
(A) AUTORADIOGRAPH
M
I
R
(B) DEGRADATION KINETICS
β -galactosidase (percent)
6–88
M-β-gal
100
I-β-gal
10
R-β-gal
β -galactosidase
1
0
30
60
90
time after adding CHX
(minutes)
Figure 6–35 Analysis of half-lives of
proteins with different N-termini (Problem
6–88). (A) Electrophoretic separation of
radioactive β-galactosidase. Antibodies
directed against β-galactosidase were
used to precipitate the protein. The bands
above β-galactosidase carry one or more
ubiquitin molecules that were attached
as a prelude to proteasomal degradation.
The N-terminal amino acids are indicated
above the lanes. (B) Disappearance of
β-galactosidases (β-gal) with time after
termination of protein synthesis by the
addition of cycloheximide (CHX). The
level of β-galactosidase is expressed as
a percentage of that present immediately
after protein synthesis was blocked.
130
Chapter 6: How Cells Read the Genome: From DNA to Protein
THE RNA WORLD AND THE ORIGINS OF LIFE
TERM TO LEARN
RNA world
DEFINITIONS
Match the deinition below with its term from the list above.
6–89
A hypothetical state of evolution that existed on Earth before modern
cells arose, in which RNA both stored genetic information and catalyzed
chemical reactions in primitive cells.
TRUE/FALSE
Decide whether the statement is true or false, and then explain why.
6–90
Protein enzymes are thought to greatly outnumber ribozymes in modern
cells because they can catalyze a much greater variety of reactions and all
of them have faster rates than any ribozyme.
THOUGHT PROBLEMS
6–91
What is so special about RNA that it is hypothesized to be an evolutionary
precursor to DNA and protein?
6–92
Discuss the following statement: “During the evolution of life on Earth,
RNA has been demoted from its glorious position as the irst replicating
catalyst. Its role now is as a mere messenger in the information low from
DNA to protein.”
6–93
Imagine a warm pond on the primordial Earth. Chance processes have
just assembled a single copy of an RNA molecule with a catalytic site that
can carry out RNA replication. his RNA molecule folds into a structure
that is capable of linking nucleotides according to instructions in an RNA
template. Given an adequate supply of nucleotides, will this single RNA
molecule be able to use itself as a template to catalyze its own replication? Why or why not?
6–94
If an RNA molecule could form a hairpin with a symmetric internal loop,
as shown in Figure 6–36, could the complement of this RNA form a similar structure? If so, would there be any regions of the two structures that
are identical? Which ones?
6–95
What is it about DNA that makes it a better material than RNA for the
storage of genetic information?
6–96
An RNA molecule with the ability to catalyze RNA replication—the linkage
of RNA nucleotides according to the information in an RNA template—
would have been a key ribozyme in the RNA world. Your advisor wants
to evolve such an RNA replicase in vitro by selection and ampliication
of rare functional molecules from a pool of random RNA sequences. he
diiculty has been to devise a strategy that models a reasonable irst step
and that can be it into an in vitro selection scheme. When you arrived
in the lab today, you found the scheme shown in Figures 6–37 and 6–38
on your desk with a note from your advisor that he wants to talk with you
about it tomorrow when he returns from an out-of-town trip. He’s even
left you some questions to “focus” the discussion.
A. As shown in Figure 6–37, the activity being selected for is the attachment
of an oligonucleotide “tag” to the end of the catalytic RNA molecule. How
is this reaction an analog of nucleotide addition to the end of a primer on
a template RNA?
B. Why is it important in the selection scheme in Figure 6–38 that the “tag”
RNA be linked to the same RNA that catalyzed its attachment?
C-U
5’-G-C-A
C-C-G
U
3’-C-G-U
G-G-C
A-C
Figure 6–36 An RNA hairpin with a
symmetric internal loop (Problem 6–94).
THE RNA WORLD AND THE ORIGINS OF LIFE
ppp
G
5’
tag
C 3’
substrate
131
N220
PP
ppp
C-G
random
sequence
3’
tag
5’
C-G
N220
3’
5’
tag
N220
3’
pool RNA
C. Why does the starting pool of RNA molecules have constant regions at
each end and a random segment in the middle? What speciic roles do
these segments play in the overall scheme?
D. How is a catalytic RNA molecule selected from the pool and speciically
ampliied?
E. Why do you suppose it is necessary to repeat the cycles of selection and
ampliication? Why not simply purify the ribozymes at the end of the irst
cycle?
Figure 6–37 Ribozyme-catalyzed
reaction selected for in this scheme for
in vitro evolution (Problem 6–96). The
random sequence in the pool RNA is
220 nucleotides in length (N220). The
3ʹ end of the substrate oligonucleotide is
complementary to the constant 5ʹ end of
the pool RNA molecules, so that it can
pair with the pool RNAs as shown.
CALCULATIONS
6–97
Curses! Your advisor has sent you an email with still more questions
about his selection scheme for evolving an RNA replicase (see Figures
6–37 and 6–38).
A. He thinks you will be able to generate about a milligram of RNA to begin
the selection. How many molecules will be present in this amount of
RNA? (Assume that an RNA nucleotide has a mass of 330 daltons and
that the RNA molecules are 300 nucleotides in length.)
B. How many diferent molecules are possible if the central 220-nucleotide
segment is completely random? What fraction of all possible molecules
will be present in your 1 mg sample?
C. If the ligation reaction could only be catalyzed by a single, unique
50-nucleotide RNA sequence, what do you suppose your chances of success would be? What does the general success of such selection schemes
imply about the range of RNA molecules that are capable of catalysis?
DATA HANDLING
Once you and your advisor had ironed out the details of the selection
scheme, the work went fairly quickly. You have now carried out 10 rounds
of selection with the results shown in Table 6–4. You have cloned and
sequenced 15 individual RNA molecules from pool 10: no two are the
pp
p
C-G
agarose
bead
tag
5’
tag
p
N220
ELUTE,
REVERSE TRANSCRIBE
N2
20
pp
5’
tag
p
N220
PCR AMPLIFY
LINKED MOLECULES
5’
C- G
N
0
22
tag
5’
tag
p
N220
T7
5’
P
6–98
PCR AMPLIFY
CATALYTIC SEGMENT,
TRANSCRIBE
next starting RNA
Figure 6–38 one round in the cyclic
selection scheme to amplify individual
ribozymes from a random pool (Problem
6–96). Each pool RNA is linked at one
end to the substrate RNA molecule and
at the other end to a complementary
DNA oligonucleotide attached to an
agarose bead for ease of manipulation.
on the right, the hairpin is not shown
for simplicity; the “p” indicates the
site of linkage between the substrate
and the presumptive catayltic RNA.
In the inal pCr ampliication, the
noncomplementary portion of the
oligonucleotide carries the promoter for
T7 rna polymerase (pT7), which allows
the inal Dna product to be transcribed
back into rna.
Chapter 6: How Cells Read the Genome: From DNA to Protein
132
TABLE 6–4 Summary of rounds of selection for a ribozyme with RNA ligation
activity (Problem 6–98).
Round
Error-prone
PCR
ligation conditions
MgCl2 (mM)
Time (hours)
0
ligation rate
(per hour)
0.000003
1
No
60
16
<0.000004
2
No
60
16
0.0008
3
No
60
16
0.0094
4
No
60
16
0.027
5
Yes
60
0.50
0.16
6
Yes
60
0.17
0.40
7
Yes
60
0.02
0.86
8
No
4
0.12
3.2
9
No
2.5
0.17
4.5
10
No
2.5
0.17
8.0
same, although 11 of the 15 have very similar sequences. Your advisor is
excited by your results and has asked you to give the next departmental
seminar. You know from your conversations with other students that you
will need to prepare careful explanations for the questions listed below.
A. Why did you use error-prone PCR, which can introduce mutations, in
some of the rounds of ampliication?
B. Why did you reduce the time and Mg2+ concentration—both changes
increase the diiculty of ligation—in successive rounds of selection?
C. How much of an improvement in ligation rate have you found in your 10
rounds of selection and ampliication?
D. Why is there still such diversity among the RNA molecules after 10 rounds
of selection and ampliication?
MCAT STYLE
Passage 1 (Questions 6–99 to 6–102)
Congenital heart defects are present in 1–2% of newborns and cause a signiicant
number of stillbirths. hey are also a major cause of heart disease in adults. Many
congenital heart defects are due to mutations in genes for transcription factors
that control expression of genes that are required for normal heart development.
For example, over 30 diferent mutations have been found in the gene for transcription factor Tbx5 in patients with congenital heart defects. To carry out its
functions, Tbx5 must bind to another transcription factor called Nkx2-5, which
also plays a role in heart development and is mutated in many individuals with
congenital heart defects. Imagine that you can isolate developing cardiac cells
with homozygous Tbx5 mutations and study the molecular defects caused by the
mutations.
6–99
In one mutant cell line, you ind that the normal, mature Tbx5 mRNA
is present at its usual levels, but that full-length Tbx5 protein cannot be
detected. What kind of mutation most likely accounts for this phenotype?
A. A mutation that blocks binding of Tbx5 to Nkx2-5
B. A mutation that causes a splicing defect
C. A mutation that disrupts the normal folding of Tbx5
D. A mutation that introduces a premature stop codon
MCAT STYLE
In another mutant cell line, you ind that levels of the mature, properly
spliced Tbx5 mRNA are greatly reduced. What kind of mutation could
account for this phenotype?
I. A short deletion in the coding region that preserves the reading frame
II. A short deletion in the coding region that disrupts the reading frame
III. A mutation in the promotor region of the Tbx5 gene
A. I
B. I and II
C. I and III
D. II and III
6–100
In another mutant cell line, you ind that full-length Tbx5 mRNA and protein are present at normal levels, yet individuals carrying this mutation
have severe defects in heart development. What kind of mutation could
cause this phenotype?
A. A mutation that blocks association of the Tbx5 mRNA with the ribosome
B. A mutation that blocks export of the Tbx5 mRNA from the nucleus
C. A mutation that causes a splicing defect
D. A mutation that disrupts binding of Tbx5 with Nkx2-5
6–101
Which one of the following is a plausible mechanism by which Tbx5
could control initiation of transcription?
A. Activation of transcription elongation factors
B. Recruitment of RNA polymerase II to promoters
C. Recruitment of splicing factors to RNA polymerase II
D. Regulation of 5ʹ cap formation
6–102
Passage 2 (Questions 6–103 to 6–105)
A key discovery in cell biology came from studying autoimmune diseases. Sera
from patients with autoimmune diseases were found to contain antibodies that
bind to cellular proteins. To characterize these proteins further, the antibodies
were used to purify the proteins they recognized. In one case, it was found that the
antibodies bound to small nuclear ribonucleoprotein complexes or snRNPs (pronounced “snurps”). he RNAs within each complex were snRNAs (small nuclear
RNAs). he functions of these snRNPs were initially mysterious, but it was eventually found that they are the key functional components of the spliceosome.
Which one of the following would have provided key evidence that
snRNPs are involved in the splicing of pre-mRNA?
A. Association of snRNPs with ribosomes
B. Base-pairing between snRNAs and pre-mRNA splice sites
C. Introns within snRNAs
D. Lariat structures within snRNAs
6–103
Autoimmune antibodies from patients were used to determine the localization of snRNPs in the cell. Which of the following best describes where
snRNP particles would be found within the cell?
A. In the cytoplasm, near ribosomes
B. In the cytoplasm, near the nucleus
C. In the nucleus, near RNA polymerase II
D. In the nucleus, near RNA polymerase III
6–104
Which one of the following proteins might be expected to be found in a
snRNP?
A. A DNA helicase
B. An ATPase
C. An enhancer protein
D. A site-speciic ribonuclease
6–105
133
Gene Regulation by the Dorsal Gradient
in the Drosophila Embryo.
This stunning image shows cross
sections through a Drosophila embryo at
about the 14th nuclear division cycle after
fertilization. The image on the left shows
the gradient of Dorsal protein as revealed
by a luorescent antibody and, on the
right, the resulting pattern of expression
of a number of genes detected by in situ
hybridization with luorescent nucleic acid
probes. at least 50 genes are controlled
by Dorsal, some activated, others
repressed.
Key to gene names: dpp,
decapentaplegic; ind, intermediate
neuroblasts defective; sog, short
gastrulation; sna, snail; vnd, ventral
neuroblasts defective.
Chapter 7
135
CHAPTER
7
Control of Gene Expression
AN OVERVIEW OF GENE CONTROL
AN OVERVIEW OF GENE CONTROL
TERMS TO LEARN
mRNA degradation control
protein activity control
RNA localization control
RNA processing control
IN THIS CHAPTER
RNA transport control
transcriptional control
translational control
DEFINITIONS
CONTROL OF TRANSCRIPTION BY
SEQUENCE-SPECIFIC
DNA-BINDING PROTEINS
TRANSCRIPTION REGULATORS
SWITCH GENES ON AND OFF
Match each deinition below with its term from the list above.
Regulates which RNAs are exported from the nucleus.
7–2
Regulates when and how often a given gene sequence is made into RNA.
7–3
Regulates which mRNA molecules are selectively destabilized in the
cytoplasm.
7–4
Regulates which mRNAs are selected to be used for protein synthesis by
ribosomes.
7–5
Regulates the splicing and modiication of RNA transcripts.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
7–6
When the nucleus of a fully diferentiated carrot cell is injected into a
frog egg whose nucleus has been removed, the injected donor nucleus is
capable of programming the recipient egg to produce a normal carrot.
7–7
he diferences in the patterns of proteins produced in diferent specialized cell types are accurately relected in the patterns of expressed
mRNAs.
MOLECULAR GENETIC
MECHANISMS THAT CREATE AND
MAINTAIN SPECIALIZED CELL
TYPES
MECHANISMS THAT REINFORCE
CELL MEMORY IN PLANTS AND
ANIMALS
POST-TRANSCRIPTIONAL
CONTROLS
REGULATION OF GENE
EXPRESSION BY NONCODING
RNAs
larger
7–1
7–8
7–9
A small portion of a two-dimensional display of proteins from human
brain is shown in Figure 7–1. hese proteins were separated on the basis
of size in one dimension and electrical charge (isoelectric point) in the
other. Not all protein spots on such displays are products of diferent
genes; some represent modiied forms of a protein that migrate to different positions. Pick out a couple of sets of spots that could represent
proteins that difer by the number of phosphates they carry. Explain the
basis for your selection.
In principle, a eukaryotic cell can regulate gene expression at any step in
the pathway from DNA to the active protein (Figure 7–2).
smaller
THOUGHT PROBLEMS
acidic
basic
Figure 7–1 Two-dimensional separation of
proteins from the human brain (Problem
7–8). The proteins were displayed using twodimensional gel electrophoresis. only a small
portion of the protein spectrum is shown.
136
Chapter 7: Control of Gene Expression
NUCLEUS
CYTOSOL
modified
protein
DNA
primary RNA
transcripts
mRNA
mRNA
Figure 7–2 Six steps at which the
pathway for eukaryotic gene expression
can be controlled (Problem 7–9).
protein
amino
acids
nucleotides
A. Place the types of control listed below at appropriate places on the diagram in Figure 7–2.
1. mRNA degradation control
2. protein activity control
3. RNA processing control
4. RNA transport and localization control
5. transcriptional control
6. translational control
B. Which of the types of control listed above are unlikely to be used in bacteria?
DATA HANDLING
7–10
In the original cloning of sheep from somatic cells, the success rate was
very low. For example, only one lamb (Dolly) was born from 277 zygotes
that were reconstructed using nuclei derived from breast cells and enucleated, unfertilized eggs. Other experiments using nuclei from embryonic or fetal lamb cells had a higher success rate, albeit still a low one.
Given the rarity of successful events, it was critical to eliminate inadvertent mating—of either the oocyte donor or the surrogate mother—as the
source of newborn lambs. To determine whether the cloned animals
were derived from the donor nuclei, the researchers analyzed DNA
microsatellites (short, repetitive DNA sequences) at four loci in surrogate
mothers and donor cells (Figure 7–3). hese loci were chosen because
many diferent lengths are present in sheep populations.
A. Do the results in Figure 7–3 argue that the lambs were derived from the
transplanted nuclei, or from an inadvertent mating? Explain your answer.
B. What would the results have looked like for the alternative you did not
choose in question A?
7–11
One of the rare examples of developmentally programmed genome
rearrangements in mammals occurs during the generation of antibody
diversity in the immune system. In B cells, which produce antibodies,
the variable (V) and constant (C) segments of immunoglobulin genes
are brought close together by deletion of a long segment of DNA that lies
1
2
3
4
lls
la
m
b
ce
lls
la
m
bs
ce
m
bs
la
SURROGATE
MOTHERS
ce
locus
lls
ORIGIN OF DONOR CELL NUCLEI
embryo
fetus
breast
Figure 7–3 Microsatellite analysis of
seven surrogate mothers, the three
different nuclear donor cell types, and
the seven lambs that were born (Problem
7–10). Four polymorphic loci were used in
the analysis. The surrogate mothers are
arranged left to right in the same order
as the lambs they gave birth to. Nuclear
donor cells were derived from embryo,
fetus, or breast. At each of the four
polymorphic loci, lanking polymerase
chain reaction (pCr) primers were used
to amplify Dna that included a particular
microsatellite. Microsatellites with
different numbers of repeats give rise to
different length pCr products.
CONTROL OF TRANSCRIPTION BY SEQUENCE-SPECIFIC DNA-BINDING PROTEINS
between them. Digestion of unrearranged germ-line DNA with a restriction nuclease that cuts in DNA lanking the V and C segments generates
two bands on a Southern blot when hybridized to radioactive probes
speciic for the V and the C segments (Figure 7–4). Would you expect the
V- and C-segment probes to hybridize to the same or diferent DNA fragments after digestion of B cell DNA with the same restriction nuclease?
Sketch a possible pattern of hybridization to B cell DNA that is consistent
with your expectations. Explain the basis for your pattern of hybridization. (Without a lot more information you cannot predict the exact pattern, so focus on the general features of the pattern.)
PROBE
137
GERM LINE
B CELLS
C
C
V
V
loading
slots
MEDICAL LINKS
7–12
Comparisons of the patterns of mRNA levels across diferent human cell
types show that the level of expression of almost every active gene is different. he patterns of mRNA abundance are so characteristic of cell type
that they can be used to determine the tissue of origin of cancer cells, even
though the cells may have metastasized to diferent parts of the body. By
deinition, however, cancer cells are diferent from their noncancerous
precursor cells. How do you suppose then that patterns of mRNA expression might be used to determine the tissue source of a human cancer?
Figure 7–4 Southern blot of DNAs from
the germ line and B cells (Problem
7–11). Germ-line DNA and B cell DNA
were digested with the same restriction
nuclease. only the hybridization to
germ-line DNA is shown.
CONTROL OF TRANSCRIPTION BY SEQUENCESPECIFIC DNA-BINDING PROTEINS
TERMS TO LEARN
cis-regulatory sequence
transcription regulator
DEFINITIONS
Match each deinition below with its term from the list above.
7–13
A protein that binds to a speciic DNA sequence and inluences the rate
at which a gene is transcribed.
7–14
A short segment of DNA, 5–10 nucleotide pairs in length, in the neighborhood of the promoter, that serves as a binding site for a protein that
modulates gene expression.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
7–15
Because the individual contacts are weak, the interactions between regulatory proteins and DNA are among the weakest in biology.
7–16
In terms of the way it interacts with DNA, the helix–loop–helix motif is
more closely related to the leucine zipper motif than it is to the helix–
turn–helix motif.
THOUGHT PROBLEMS
7–17
Figure 7–5 shows a short stretch of a DNA helix displayed as a spaceilling model. Indicate the major and minor grooves and provide a length
scale. Is it possible to tell the polarity of each of the strands in this igure?
7–18
Explain how DNA-binding proteins can make sequence-speciic contacts to a double-stranded DNA molecule without breaking the hydrogen bonds that hold the bases together. Indicate how, by making such
Figure 7–5 A space-illing model of a Dna
duplex (Problem 7–17).
138
Chapter 7: Control of Gene Expression
MAJOR GROOVE
MAJOR GROOVE
H
H
N H
O
N
H
N
O
H
N
H
H
H
N
N
N
N
cytosine
H
H
H
H N
O
C
H
N
H
N
O
H
guanine
H
thymine
MINOR GROOVE
N
N
N
H
N
adenine
MINOR GROOVE
contacts, a protein can distinguish a C-G from a T-A base pair. Use
Figure 7–6 to indicate what sorts of noncovalent bonds (hydrogen bonds,
electrostatic attractions, or hydrophobic interactions) could be used to
discriminate between C-G and T-A. (You do not need to specify any particular amino acid on the protein.)
7–19
What are the two fundamental components of a genetic switch?
7–20
he nucleus of a eukaryotic cell is much larger than a bacterium, and it
contains much more DNA. As a consequence, a transcription regulator
in a eukaryotic cell must be able to select its speciic binding site from
among many more unrelated sequences than does a transcription regulator in a bacterium. Does this present any special problems for eukaryotic gene regulation?
Consider the following situation. Assume that the eukaryotic nucleus
and the bacterial cell each have a single copy of the same DNA binding
site. In addition, assume that the nucleus is 500 times the volume of the
bacterium, and has 500 times as much DNA. If the concentration of the
transcription regulator that binds the site were the same in the nucleus
and in the bacterium, would the regulator occupy its binding site equally
as well in the eukaryotic nucleus as it does in the bacterium? Explain
your answer.
7–21
One type of zinc inger motif consists of an α helix and a β sheet held
together by a zinc ion (Figure 7–7). When this motif binds to DNA, the α
helix is positioned in the major groove, where it makes speciic contacts
with the bases. his type of zinc inger is often found in a cluster with
additional zinc ingers, an arrangement that allows strong and speciic
DNA–protein interactions to be built up through a repeating basic structural unit. Why do you suppose this motif is thought to enjoy a particular
advantage over other DNA-binding motifs when the strength and speciicity of the DNA–protein interaction needs to be adjusted during evolution?
7–22
Many transcription regulators form dimers of identical or slightly diferent subunits on the DNA. Suggest two advantages of dimerization.
7–23
Genetic analyses in bacteria in the 1950s provided the irst evidence
for the existence of transcription regulators. he lambda repressor, one
such regulator, is encoded by the bacterial virus, bacteriophage lambda.
he lambda repressor binds as a dimer to critical sites on the bacteriophage lambda genome to keep the lytic genes turned of, which allows
the bacteriophage lambda genome to be maintained as a silent resident
in the bacterial genome. Each molecule of the repressor consists of an
N-terminal DNA-binding domain and a C-terminal dimerization domain
(Figure 7–8). Upon induction (for example, by irradiation with ultraviolet light), the genes for lytic growth are expressed, bacteriophage lambda
Figure 7–6 C-G and T-A base pairs
(Problem 7–18).
C
H 24
C7
H 20
Zn
C4
N
Figure 7–7 Zinc inger Dna-binding motif
(Problem 7–21). The zinc ion interacts with
Cys (C) and his (h) side chains so that
the α helix is held tightly to one end of the
β sheet.
CONTROL OF TRANSCRIPTION BY SEQUENCE-SPECIFIC DNA-BINDING PROTEINS
C
C
C
C
C
C
cleavage
site
+
N
N
repressor monomers
N
N
N
repressor dimer
N
DNA-binding site
progeny are produced, and the bacterial cell lyses to release the viral
progeny. Induction is initiated by cleavage of the lambda repressor at a
site between the DNA-binding domain and the dimerization domain. In
the absence of bound repressor, RNA polymerase initiates transcription
of the lytic genes, triggering lytic growth. Given that the number (concentration) of DNA-binding domains is unchanged by cleavage of the repressor, why do you suppose its cleavage results in its removal from the DNA?
7–24
he diferentiation of muscle cells from the somites of the developing
embryo is controlled by myogenin, a member of the MyoD family of
helix–loop–helix transcription regulators. Myogenin functions as a heterodimer with another member of the MyoD family of helix–loop–helix
proteins (Figure 7–9A). he activity of myogenin must be carefully controlled lest it trigger premature expression of the muscle program of
cell diferentiation. he myogenin gene is turned on in advance of the
time when it is needed, but myogenin is prevented from functioning
by phosphorylation of its DNA-binding domain and by its tight binding to Id, a helix–loop–helix protein that lacks a DNA-binding domain
(Figure 7–9B). Explain how phosphorylation of the DNA-binding domain
and dimerization with Id might act to keep myogenin nonfunctional.
CALCULATIONS
7–25
One method for determining the DNA sites bound by a transcription
regulator is to mix random sequences of DNA with the regulator, separate the bound sequences from the starting mixture, amplify the bound
sequences by PCR, and then repeat this binding-release-ampliication
cycle until a consensus sequence emerges. But is it reasonable to expect
that all possible consensus-site binding sequences will actually be present in the starting sample of oligonucleotides?
(A) MyoD–HLH HETERODIMER
BOUND TO DNA
HLH
(B) PHOSPHO–MyoD/Id
HETERODIMER
MyoD
Id
MyoD
P
Figure 7–9 The transcription regulator myogenin (Problem 7–24). (A) Myogenin as
part of a heterodimer bound to DNA. HlH stands for helix–loop–helix. (B) An inactive
form of phosphorylated myogenin bound to Id.
139
Figure 7–8 Domains of the lambda
repressor and the binding of repressor
dimers to DNA (Problem 7–23).
140
Chapter 7: Control of Gene Expression
TABLE 7–1 Nucleotide sequences of selected and amplified DNAs
(Problem 7–25).
GAATTCGCCTCGAGCACATCATTGCCCATATATGGCACGACAGGATCC
GAATTCGCCTCTTCTAATGCCCATATATGGACTTGCTCGACAGGATCC
GGATCCTGTCGGTCCTTTATGCCCATATATGGTCATTGAGGCGAATTC
GAATTCGCCTCATGCCCATATATGGCAATAGGTGTTTCGACAGGATCC
GAATTCGCCTCTATGCCCATATAAGGCGCCACTACCCCGACAGGATCC
GAATTCGCCTCGTTCCCAGTATGCCCATATATGGACACGACAGGATCC
GGATCCTGTCGACACCATGCCCATATTTGGTATGCTCGAGGCGAATTC
GAATTCGCCTCATTTATGAACATGCCCTTATAAGGACCGACAGGATCC
GAATTCGCCTCTAATACTGCAATGCCCAAATAAGGAGCGACAGGATCC
GAATTCGCCTCATGCCCAAATATGGTCATCACCTACACGACAGGATCC
Underlined sequences correspond to PCR primer sites. The sequences have been
ordered so that the binding sites are all oriented in the same direction (to make it
easier to pick out the consensus sequence by eye).
Consider the following speciic example. You wish to test all possible consensus sequences that are 14 nucleotides long. You synthesize a
population of oligonucleotides that carry a central 26-nucleotide-long
random sequence, lanked on either side by 25-nucleotide-long deined
sequences to serve as primer sites for PCR ampliication. You convert the
single-stranded oligonucleotides to double-stranded ones, using one of
the PCR primers. You begin the irst cycle of binding with a 0.4 ng sample
of the synthesized population of double-stranded oligonucleotides.
A. How many double-stranded oligonucleotides, 76-nucleotide pairs long,
are present in the 0.4 ng sample with which you started the experiment?
(he average mass of a nucleotide pair is 660 daltons.)
B. Assuming that the starting population of oligonucleotides was truly random in the 26 central nucleotides, would you expect to ind all possible
14-base-pair-long sequences represented in the starting sample?
C. he sequences of 10 individual clones resulting from this procedure are
shown in Table 7–1. What is the consensus sequence to which the transcription regulator binds?
DATA HANDLING
When Jacob and Wollman tried to check the genetic linkage between
the Gal gene and an integrated bacteriophage lambda genome (termed
a lambda prophage), they discovered a surprising phenomenon they
referred to as “erotic induction” (which was later called zygotic induction
for publication). In a bacterial mating used to determine genetic linkage,
a portion of the chromosome is transferred via a narrow tube from the
donor bacterium to the recipient. Jacob and Wollman found that if the
donated chromosome carried a lambda prophage, but the recipient cell
did not, lambda growth was induced in the recipient cell, which then
lysed, producing lambda phage. If the recipient cell carried the lambda
prophage, however, no lysis was observed. A summary of results from all
their matings is shown in Figure 7–10.
A. Are these results consistent with the notion that a transcription regulator
encoded by the prophage normally represses the bacteriophage’s lytic
genes. Why or why not?
recipient
lambda– lambda+
donor
lambda+ lambda–
7–26
no
lysis
no
lysis
lysis +
lambda
phage
no
lysis
Figure 7–10 Results of matings between
bacteria with and without lambda
prophages (Problem 7–26). lambda–
indicates the absence of a prophage;
lambda+ indicates its presence.
CONTROL OF TRANSCRIPTION BY SEQUENCE-SPECIFIC DNA-BINDING PROTEINS
B. Suppose that the prophage prevented lytic growth by expressing a transcription regulator that turned on a gene for an anti-lysis protein. Would
the results of the matings have been the same or diferent? Explain your
answer.
7–27
7–28
Transcription regulators often ind their speciic sites much faster than
would be anticipated by simple three-dimensional difusion. he Lac
repressor, for example, associates with the Lac operator—its DNA binding site—more than 100 times faster than expected for three-dimensional
difusion. Clearly, the repressor must ind the operator by mechanisms
that reduce the dimensionality or volume of the search in order to hasten
target acquisition.
Several techniques have been used to investigate this problem. One
of the most elegant used luorescent RNA polymerase molecules that
could be followed individually. An array of DNA molecules was aligned
in parallel by an electrophoretic technique and anchored to a glass slide.
Fluorescent RNA polymerase molecules were then allowed to low across
them at an oblique angle (Figure 7–11A). Traces of individual RNA polymerases showed that about half lowed in the same direction as the bulk
and about half deviated from the bulk low in a characteristic manner
(Figure 7–11B). If the RNA polymerase molecules were irst incubated
with short DNA fragments containing a strong promoter, all the traces
followed the bulk low.
A. Ofer an explanation for why some RNA polymerase molecules deviated
from the bulk low as shown in Figure 7–11B. Why did incubation with
short DNA fragments containing a strong promoter eliminate traces that
deviated from the bulk low?
B. Do these results suggest an explanation for how transcription regulators
manage to ind their sites faster than expected by difusion?
C. Based on your explanation, would you expect a transcription regulator
to ind its target site faster in a population of short DNA molecules or in
a population of long DNA molecules? Assume that the concentration of
DNA is the same in both populations and that both populations have the
same number of target sites.
he binding of a transcription regulator to a DNA sequence can cause
the DNA to bend to make appropriate contacts with groups on the surface of the protein. Such protein-induced DNA bending can be readily
detected by the way the protein–DNA complexes migrate through polyacrylamide gels. he rate of migration of bent DNA depends on the average distance between its ends as it gyrates in solution: the more bent the
DNA, the closer together the ends are on average, and the more slowly
it migrates. If there are two sites of bending, the end-to-end distance
depends on whether the bends are in the same (cis) or opposite (trans)
directions (Figure 7–12A).
You have shown that the catabolite activator protein (CAP) causes
DNA to bend by more than 90° when it binds to its regulatory site. You
wish to know the details of the bent structure. Speciically, is the DNA
at the center of the CAP-binding site bent so that the minor groove of
the DNA helix is on the inside, or is it bent so that the major groove is
on the inside? To answer this, you prepare two kinds of constructs, as
shown in Figure 7–12B. In one, you place two CAP-binding sequences
on either side of a central site into which you insert a series of DNA segments that vary from 10 to 20 nucleotide pairs in length. In the other, you
lank the insertion site with one CAP-binding sequence and one (A5N5)4
sequence, which is known to bend with the major groove on the inside.
You measure the migration of the CAP-bound constructs relative to the
corresponding CAP-bound DNA with no insert. You then plot the relative
migration versus the number of nucleotide pairs between the centers of
bending (Figure 7–12C and D).
141
(A) EXPERIMENTAL SET-UP
bulk flow of
RNA polymerase
molecules
aligned
DNA
molecules
(B) SINGLE RNA POLYMERASE MOLECULES
Figure 7–11 Interactions of individual
RNA polymerase molecules with DNA
(Problem 7–27). (A) Experimental set-up.
DNA molecules are aligned and anchored
to a glass slide at their ends, and highly
luorescent rna polymerase molecules
are allowed to low across them.
(B) Traces of two individual rna
polymerase molecules. The one on the
left has traveled with the bulk low, and
the one on the right has deviated from it.
The scale bar is 10 μm.
142
Chapter 7: Control of Gene Expression
(A) DNA BENDING BY CAP
(B) TWO BENDY CONSTRUCTS
CAP
insert
CAP
cis
centers of bending
(A5N5)4
trans
insert
CAP
centers of bending
relative migration
(C) CAP–CAP
(D) (A5N5)4–CAP
1.2
1.2
1.1
1.1
1.0
1.0
0.9
0.9
0.8
0.8
77
79
81
83
85
87
nucleotides between
centers of bending
89
93
95
97
99
101 103 105
nucleotides between
centers of bending
A. Assuming that there are 10.6 nucleotides per turn of the DNA helix, estimate the number of turns that separate the centers of bending of the
two CAP-binding sites at the point of minimum relative migration. How
many helical turns separate the centers of bending at the point of maximum relative migration?
B. Is the relationship between the relative migration and the separation of
the centers of bending of the CAP sites what you would expect, assuming
that the cis coniguration migrates slower than the trans coniguration
(Figure 7–12C)? Explain your answer.
C. How many helical turns separate the centers of bending at the point of
minimum migration of the construct with one CAP site and one (A5N5)4
site (Figure 7–12D)?
D. Which groove of the helix faces the inside of the bend at the center of
bending of the CAP site?
7–29
he Fos and Jun oncogenes encode proteins that form a heterodimeric
transcription regulator. Leucine zipper domains in each protein mediate
their dimerization through coiled-coil interactions. Dimerization juxtaposes the DNA-binding domains of each protein, positioning them for
interaction with regulatory sites in DNA. he dynamics of Fos–Jun interaction in the presence and absence of AP-1 DNA (the sequence to which
the Fos–Jun heterodimer binds) were investigated by luorescence resonance energy transfer (FRET), which is well suited for the rapid measurements that are necessary in such studies.
Fos was tagged with luorescein (Fos–F) and Jun was tagged with
rhodamine (Jun–R), as shown in Figure 7–13A. Fluorescein absorbs
light at 490 nm and emits light at 530 nm, whereas rhodamine absorbs
light at 530 nm and emits light at 603 nm. When Fos–F and Jun–R are
brought into close proximity through heterodimerization, light energy
absorbed by luorescein at 490 nm is eiciently transferred to rhodamine through nonradiative energy transfer and emitted by rhodamine at
603 nm. Dimerization thus decreases luorescence by luorescein at
530 nm and increases luorescence by rhodamine at 603 nm, as shown in
Figure 7–13B. In the presence of AP-1 DNA, FRET is even more eicient
Figure 7–12 Bending of DNA by CAP
binding (Problem 7–28). (A) The cis and
trans conigurations of a pair of bends.
(B) Two constructs used to investigate
Dna bending by Cap binding.
(C) relative migration as a function of
the number of nucleotides between the
centers of bending in the Cap–Cap
construct. (D) relative migration as a
function of the number of nucleotides
between the centers of bending in the
(a5n5)4–Cap construct.
CONTROL OF TRANSCRIPTION BY SEQUENCE-SPECIFIC DNA-BINDING PROTEINS
rhodamine
fluorescein
(B) EMISSION SPECTRA
(C) EXCHANGE KINETICS
fluorescein
530 nm
Jun–R
AP-1 DNA
Fos–F
relative fluorescence
1.0
Fos–F
0.8
0.6
Fos–F
+ Jun–R
rhodamine
603 nm
0.4
Fos–F + Jun–R
+ AP-1 DNA
0.2
0
500
relative fluorescence at 603 nm
(A)
143
+ AP-1 DNA
1.0
+ Fos
0.8
0.6
0.4
− AP-1 DNA
0.2
0
540
580
620
wavelength (nm)
660
0
200
400 600 800 1000
time (seconds)
Figure 7–13 Dynamics of Fos–Jun heterodimerization in the presence and absence of AP-1 DNA (Problem 7–29).
(A) Arrangement of luorophores in Fos–F and Jun–r. In the dimer, the excited luorescein on Fos transfers energy to the
rhodamine on Jun. (B) emission spectra of Fos–F alone, Fos–F with Jun–r, and Fos–F with Jun–r and ap-1 Dna. The
emission spectra were recorded between 500 and 700 nm after excitation at 490 nm. (C) analysis of the exchange of
Fos–F and Jun–r in the presence and absence of ap-1 Dna. rhodamine luorescence at 603 nm was followed over time
after excitation of luorescein at 490 nm. a 10-fold excess of unlabeled Fos was added at the time indicated by the arrow.
(Figure 7–13B), indicating that binding to the DNA brings the two luorophores into even closer proximity.
FRET was also used to measure the ability of Fos–Jun dimers to
exchange subunits with monomers in solution in the presence and
absence of AP-1 DNA (Figure 7–13C). Fos–F and Jun–R were preincubated in the absence of DNA to allow free heterodimers to form, or in
its presence to allow DNA-bound heterodimers to form. A 10-fold excess
of Fos (without luorescein) was then added to both solutions, and rhodamine luorescence at 603 nm was followed after excitation at 490 nm
(Figure 7–13C).
A. Do free heterodimers exchange subunits with added unlabeled Fos?
Do heterodimers bound to DNA exchange subunits? How can you tell?
Explain any signiicant diferences in the behavior of free and DNAbound heterodimers.
B. In most cells, there are many distinct leucine zipper transcription regulators, several of which can interact to form a variety of heterodimers. If the
results in Figure 7–13C were typical of the leucine zipper proteins, what
would they imply about the leucine zipper heterodimers in cells?
7–30
he binding of a protein to DNA can protect it from chemical cleavage,
which is the basis for the technique known as DNA footprinting. You
have used DNA footprinting to determine the DNA binding site of a
transcription regulator after labeling one strand. To check your results,
you repeat the experiment after labeling the other strand of the duplex.
You ind that the footprints are slightly ofset from one another relative
to the sequence of the DNA (Figure 7–14). If the protein binds to the
same duplex in both cases, how can the footprints on the two strands be
diferent?
top
strand
top
strand
no protein
5ʹ-CTGTGTGTATGCTGGGAAGGACTT
+ protein
bottom
strand
+ protein
GACACACATACGACCCTTCCTGAA– 5ʹ
Figure 7–14 Dna footprints (Problem
7–30). The sequences of the Dna from
the top and bottom strands around the
footprint are shown. The 5ʹ ends of the
top or bottom strands were labeled with
32p. Chemical cleavage was used to
introduce Dna breaks, which are fairly
randomly distributed, as shown by the
roughly equal intensities of the individual
bands. each band corresponds to a
fragment of Dna that differs from those
on either side by one nucleotide.
144
Chapter 7: Control of Gene Expression
TRANSCRIPTION REGULATORS AS GENE SWITCHES
TERMS TO LEARN
cis-regulatory sequence
gene
gene control region
promoter
DEFINITIONS
Match each deinition below with its term from the list above.
7–31
Nucleotide sequence in DNA to which RNA polymerase binds to begin
transcription.
7–32
he whole expanse of DNA involved in regulating and initiating transcription of a gene.
7–33
he segment of DNA that is transcribed into RNA.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
7–34
Many transcription regulators in eukaryotes can act even when they are
bound to DNA thousands of nucleotide pairs away from the promoter
they inluence.
7–35
he close-packed arrangement of bacterial genes and genetic switches
may have developed from more extended forms of switches in response
to the evolutionary pressure to maintain a small genome.
THOUGHT PROBLEMS
7–36
he genes encoding the enzymes for arginine biosynthesis are located
at several positions around the genome of E. coli. he ArgR transcription
regulator coordinates their expression. he activity of ArgR is modulated
by arginine. Upon binding arginine, ArgR dramatically changes its ainity for the cis-regulatory sequences in the promoters of the genes for the
arginine biosynthetic enzymes. Given that ArgR is a transcription repressor, would you expect that ArgR would bind more tightly, or less tightly, to
the regulatory sequences when arginine is abundant? If ArgR functioned
instead as a transcription activator, would you expect the binding of arginine to increase, or to decrease, its ainity for its regulatory sequences?
Explain your answers.
7–37
Bacterial cells can take up the amino acid tryptophan from their surroundings, or, if the external supply is insuicient, they can synthesize
tryptophan from small molecules in the cell. he tryptophan repressor inhibits transcription of the genes in the tryptophan operon, which
encodes the tryptophan biosynthetic enzymes. Upon binding tryptophan, the tryptophan repressor binds to a site in the promoter of the
operon.
A. Why is tryptophan-dependent binding to the operon a useful property
for the tryptophan repressor?
B. What would you expect to happen to the regulation of the tryptophan
biosynthetic enzymes in cells that express a mutant form of the tryptophan repressor that (i) cannot bind to DNA or (ii) binds to DNA even
when no tryptophan is bound to it?
C. What would happen in scenarios (i) and (ii) if the cell produced normal
tryptophan repressor from a second, unmutated copy of the gene?
7–38
In Figure 7–15, the bacterial activator protein CAP and the Lac repressor have been placed in the four possible combinations on their binding
TRANSCRIPTION REGULATORS AS GENE SWITCHES
145
GLUCOSE LACTOSE
Figure 7–15 Arrangement of binding sites
and the four possible combinations of
transcription regulators on the promoter
for the Lac operon (Problem 7–38).
OPERON ACTIVITY
Lac repressor
CAP
Lac repressor
CAP
RNA polymerase
LacZ
sites in the promoter for the Lac operon. Each combination of transcription regulators corresponds to the expected binding in a particular mixture of glucose and lactose. For each of the four combinations, indicate
on the left-hand side of the igure which sugars are present and, on the
right-hand side, whether the operon is expected to be turned on or of.
7–39
Imagine that you have created a fusion between the Trp operon,
which encodes the enzymes for tryptophan biosynthesis, and the Lac
operon, which encodes the enzymes necessary for lactose utilization (Figure 7–16). Under which set of conditions (A–F below) will
β-galactosidase be expressed in the strain that carries the fused operon?
A. Only when lactose and glucose are both absent.
B. Only when lactose and glucose are both present.
C. Only when lactose is absent and glucose is present.
D. Only when lactose is present and glucose is absent.
E. Only when tryptophan is absent.
F. Only when tryptophan is present.
7–40
When cis-regulatory sequences were initially found to inluence activity at distant promoters, two principal models were invoked to explain
this action at a distance. In the “DNA looping” model, direct interactions
between transcription regulators bound at cis-regulatory sequences and
the distant promoters were proposed to stimulate RNA polymerase. In
the “scanning” model, RNA polymerase (or a transcription regulator)
was proposed to bind at the regulatory sequence and then slide along the
DNA until it reached the promoter. hese two models were distinguished
using a cis-regulatory sequence on one piece of DNA and the β-globin
gene with its promoter on a separate piece of DNA (Figure 7–17). he
β-globin gene was not expressed from the mixture of pieces. However,
when the two segments of DNA were joined via a protein linker, the
β-globin gene was expressed.
Trp
regulatory
region
Lac
regulatory
region
Trp operon
E
D
C
Figure 7–16 Separated (normal) and fused
Trp and Lac operons (Problem 7–39).
B
A
Lac operon
z
encodes
β-galactosidase
FUSION
Trp
regulatory
region
Trp–Lac fusion operon
E
D
C
z
encodes
β-galactosidase
y
a
y
a
146
Chapter 7: Control of Gene Expression
biotin
cis-regulatory
sequence
β-globin
+ avidin
cis-regulatory
sequence
β-globin
How does this experiment distinguish between the DNA looping
model and the scanning model? Explain your answer.
7–41
Some transcription regulators bind to DNA and cause the double helix to
bend at a sharp angle. Such “bending proteins” can afect the initiation
of transcription without directly contacting any other protein. Can you
devise a plausible explanation for how such proteins might work to modulate transcription? Draw a diagram that illustrates your explanation.
7–42
he yeast Gal4 transcription activator comprises two domains: a DNAbinding domain and an activation domain. he DNA-binding domain
allows Gal4 to bind to appropriate DNA sequences located near genes
that are required for metabolism of the sugar galactose. he activation
domain binds to components of the transcriptional machinery (including RNA polymerase), attracting them to the promoter, so the regulated
genes can be turned on. In the absence of Gal4, the galactose genes cannot be turned on. When Gal4 is expressed normally, the genes can be
maximally activated. When Gal4 is massively overexpressed, however,
the galactose genes are turned of. Why do you suppose that too much
Gal4 squelches expression of the galactose genes?
7–43
How are histone modiication enzymes and chromatin remodeling complexes recruited to unmodiied chromatin, and how are they thought to
aid in the activation of transcription from previously silent genes?
7–44
How is it that protein–protein interactions that are too weak to cause proteins to assemble in solution can nevertheless allow the same proteins to
assemble into complexes on DNA?
7–45
Consider the following argument: “If the expression of every gene
depends on a set of transcription regulators, then the expression of
these transcription regulators must also depend on the expression of
other transcription regulators, and their expression must depend on the
expression of still other transcription regulators, and so on. Cells would
therefore need an ininite number of genes, most of which would code
for transcription regulators.” How does the cell get by without having to
achieve the impossible?
DATA HANDLING
7–46
E. coli proliferates faster on the monosaccharide glucose than it does on
the disaccharide lactose for two reasons: (1) lactose is taken up more
slowly than glucose and (2) lactose must be hydrolyzed to glucose and
galactose (by β-galactosidase) before it can be further metabolized.
When E. coli is grown on a medium containing a mixture of glucose
and lactose, it proliferates with complex kinetics (Figure 7–18, squares).
he bacteria proliferate faster at the beginning than at the end, and
there is a lag between these two phases when they virtually stop dividing. Assays of the concentrations of the two sugars in the medium show
Figure 7–17 Stimulation of β-globin gene
expression by a cis-regulatory sequence
linked via a protein bridge (Problem
7–40). Each DNA molecule carries biotin
attached to one end, as shown. In the
presence of the protein avidin, the
two molecules are linked together and
transcription occurs, as shown by the
arrow above the β-globin gene.
TRANSCRIPTION REGULATORS AS GENE SWITCHES
147
Figure 7–18 Proliferation of E. coli
on a mixture of glucose and lactose
(Problem 7–46).
bacteria per mL
108
10
glucose
bacteria
107
5
106
0
0
100
glucose concentration (mM)
β−galactosidase (arbitrary units)
β-galactosidase
200
time (minutes)
that glucose falls to very low levels after a few cell doublings (Figure 7–18,
circles), but lactose remains high until near the end of the experimental
time course (not shown). Although the concentration of lactose is high
throughout most of the experiment, β-galactosidase, which is regulated
as part of the Lac operon, is not induced until more than 100 minutes
have passed (Figure 7–18, triangles).
A. Explain the kinetics of bacterial proliferation during the experiment.
Account for the rapid initial rate, the slower inal rate, and the delay in
the middle of the experiment.
B. Explain why the Lac operon is not induced by lactose during the rapid
initial phase of bacterial proliferation.
7–47
7–48
Transcription of the bacterial gene encoding the enzyme glutamine
synthetase is regulated by the availability of nitrogen in the cell. he key
transcription regulator is the NtrC protein, which stimulates transcription when it is phosphorylated. Transcription of the gene for glutamine
synthetase can be achieved in vitro by adding RNA polymerase, a special
sigma factor, and phosphorylated NtrC. Although NtrC binds regardless
of its phosphorylation state and RNA polymerase binds strongly to the
promoter in the absence of NtrC, transcription is absolutely dependent
on phosphorylated NtrC.
Activation of transcription by NtrC was characterized using three different templates: the normal gene with ive regulatory sequences, a gene
with all of the NtrC-binding sites deleted, and a gene with three NtrCbinding sites at its 3ʹ end (Figure 7–19). For half-maximal rates of transcription, the normal gene (Figure 7–19A) required 5 nM NtrC, the gene
with 3ʹ NtrC-binding sites (Figure 7–19C) required 10 nM NtrC, and the
gene without NtrC-binding sites (Figure 7–19B) required 50 nM NtrC.
A. If RNA polymerase can bind to the promoter of the glutamine synthetase
gene in the absence of NtrC, why do you suppose NtrC is needed to activate transcription?
B. If NtrC can bind to its binding sites regardless of its state of phosphorylation, why do you suppose phosphorylation is necessary for transcription?
C. If NtrC can activate transcription even when its binding sites are absent,
what role do the binding sites play?
Coactivators do not bind to DNA, but instead serve as bridging molecules that link transcription activators to general transcription factors
at the promoter. he SAGA complex in yeast is a large multiprotein complex that is required for transcription of many genes. SAGA contains a
variety of proteins, including a histone acetyl transferase and a subset of
TATA-binding protein associated factors (TAFs). If SAGA functions as a
(A)
glutamine synthetase
NtrCgene
binding sites
promoter
(B)
glutamine synthetase
gene
promoter
(C) glutamine synthetase
gene
NtrCbinding sites
promoter
Figure 7–19 Three templates for studying
the role of NtrC in transcription of the
glutamine synthetase gene (Problem
7–47). (A) The normal gene with intact
upstream regulatory sequences. (B) A
gene with all of the NtrC-binding sites
deleted. (C) A gene with three NtrCbinding sites at the 3ʹ end of the gene.
Chapter 7: Control of Gene Expression
148
(A) MAP OF THE Gal1-Gal10 LOCUS
Gal10
Gal1
UAS
A
B
C
D
E
F
G
(B) PCR PRODUCTS FROM CHROMATIN IMMUNOPRECIPITATES
Spt3
Spt20
6
4
Gal/Raf
Gal/Raf
8
6
4
2
2
0
IN
A
B
C
D
E
F
G
IN
0
A
B
C
D
E
F
G
Gal
Raf
coactivator, it should be physically present at the promoters it regulates
under appropriate conditions.
To test this idea, you focus on the regulation of the Gal1 and Gal10
genes by the Gal4 activator, which binds to cis-regulatory sequences
(termed UAS) adjacent to their promoters (Figure 7–20A, green boxes).
When cells are grown on galactose, the Gal1 and Gal10 genes are transcribed; when the cells are grown on the sugar rainose, the genes are
not transcribed.
To determine whether the SAGA complex associates with Gal4 on the
chromosome, you use the technique of chromatin immunoprecipitation. You shear the chromatin to small pieces and precipitate those that
are bound to the SAGA complex, using antibodies against two components of the complex—Spt3 and Spt20. You strip the precipitated chromatin of protein and analyze the DNA by quantitative PCR to measure
the amounts of speciic DNA segments in the precipitates. he ratio of
precipitated DNA sequences from galactose-grown cells to those from
rainose-grown cells (Gal/Raf ) is shown for several segments around the
galactose promoter in Figure 7–20B.
A. Which of the tested fragments would you have expected to be enriched if
SAGA behaved as a coactivator? Explain your answer.
B. Does SAGA meet the criteria for a coactivator? Is it physically present at
the promoter under appropriate conditions?
7–49
Genes expressed at the yeast mating type (Mat) locus on chromosome
III determine the mating type of a haploid yeast cell. Identical matingtype genes also exist at two other loci—Hml and Hmr—on the same chromosome (Figure 7–21A). At these two silent loci, the mating-type genes
are not expressed, even though all the signals for expression are present.
Each silent mating-type locus is bracketed by two short DNA sequences,
designated E and I, which bind a variety of proteins and serve to establish
and maintain repression of genes within each locus.
To investigate the function of these elements, you insert the Ura3
gene at diferent locations between E and I and outside the Hml locus,
as shown in Figure 7–21B. You then test the growth of these strains in
complete medium, in the absence of uracil (–uracil), and in the presence
of 5-luoroorotic acid (+FOA) (Figure 7–21B). hese growth conditions
Figure 7–20 Analysis of SAGA
association with the Gal promoter
(Problem 7–48). (A) organization of
the Gal10 and Gal1 genes. A common
UAS serves both promoters, which
are indicated by green boxes. The
positions of the PCR fragments that
were used for quantiication are shown
by a–G below. (B) results of chromatin
immunoprecipitation using antibodies
against spt3 or spt20. The strains
were grown in medium containing
rafinose (raf) or galactose (Gal),
their chromatin was fragmented and
immunoprecipitated, and the Dna was
ampliied by quantitative pCr. The pCr
products from strains grown on rafinose
were loaded onto the gels and run for a
few minutes before the pCr products
from strains grown on galactose, so that
the products could be compared within
the same lanes. Finally, the amounts
of pCr products were measured,
normalized to the input chromatin before
immunoprecipitation (In), and the ratio
(Gal/raf) was calculated. about 50- to
100-fold less input chromatin (In) was
loaded per lane than immunoprecipitated
chromatin, relecting the ineficiency of
chromatin immunoprecipitation.
MOLECULAR GENETIC MECHANISMS THAT CREATE AND MAINTAIN SPECIALIZED CELL TYPES
149
(A) YEAST SILENT MATING-TYPE LOCI
α2 α1
E
a2 a1
centromere
I
Hml
E
Hmr
I
3 kb
(B) INSERTION OF Ura3 GENES NEAR Hml
0
E
A
Hml
B
I
C
D
E
F
complete medium
– uracil
0
1
1 Ura3
2
Ura3
2
3
Ura3
3
4 Ura3
4
Ura3
5
6
Ura3
6
7
Ura3
5
7
8 Ura3
8
Ura3
9
10
9
Ura3
test for the activity of the Ura3 gene: in complete medium, expression of
the Ura3 protein is irrelevant; in the absence of uracil, expression of Ura3
is required for growth; and in the presence of FOA, expression of Ura3 is
lethal to the cell.
Do these control elements speciically repress the mating-type genes,
or do they act like insulators that control the expression of any gene
placed between them?
MOLECULAR GENETIC MECHANISMS THAT CREATE
AND MAINTAIN SPECIALIZED CELL TYPES
TERMS TO LEARN
cell memory
+ FOA
induced pluripotent stem (iPS) cell
DEFINITIONS
Match each deinition below with its term from the list above.
7–50
Cell derived from a ibroblast, by artiicial expression of speciic transcription regulators, that looks and behaves like an embryonic stem cell.
7–51
he property that allows a proliferating cell to maintain its identity
through subsequent cell divisions.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
7–52
In most organisms, extracellular molecules that bind to receptors on
the cell surface communicate the positional cues that allow cells in the
embryo to diferentiate into appropriate cell types.
7–53
he ibroblasts that are converted to muscle cells by expression of the
transcription regulator MyoD have probably already accumulated a
number of transcription regulators that can cooperate with MyoD to
switch on muscle-speciic genes.
10
Figure 7–21 Effects of E and I elements
at Hml on the expression of the Ura3
gene inserted nearby (Problem 7–49).
(A) Arrangement of E and I elements
around Hml and Hmr. The mating-type
genes are shown as α1, α2, a1, and a2.
(B) locations of the Ura3 gene inserted
around the Hml locus. orientations
of the Ura3 gene are indicated by the
direction of lettering. Growth phenotypes
of each strain are indicated in the panel
on the right. Strain 0 is the parental
strain with no Ura3 gene; strains 1 to
10 have the Ura3 gene inserted as
indicated. Serial dilutions of each strain
were spotted on agar plates (with the
highest concentration on the left). The
agar contained complete medium,
complete medium minus uracil (–uracil),
or complete medium plus FoA (+FoA).
White areas indicate growth.
150
Chapter 7: Control of Gene Expression
7–54
Once cells have diferentiated to their inal specialized forms, they never
again alter expression of their genes.
THOUGHT PROBLEMS
7–55
(A) PROPHAGE STATE
cI gene
Bacteriophage lambda can replicate as a prophage or lytically. In the
prophage state, the viral DNA is integrated into the bacterial chromosome and is copied once per cell division. In the lytic state, the viral DNA
is released from the chromosome and replicates many times. his viral
DNA then produces viral coat proteins that enclose the replicated viral
genomes to form many new virus particles, which are released when the
bacterial cell bursts.
hese two states are controlled by the transcription regulators cI
and Cro, which are encoded by the virus. In the prophage state, cI is
expressed; in the lytic state, Cro is expressed. In addition to regulating
the expression of other genes, cI is a repressor of transcription of the
gene that encodes Cro, and Cro is a repressor of the gene that encodes cI
(Figure 7–22).
When bacteria containing a lambda prophage are briely irradiated
with ultraviolet (UV) light, cI protein is degraded.
A. What will happen next?
B. Will the change in question A be reversed when the UV light is switched
of?
C. How is this mechanism beneicial to the virus?
7–56
Imagine the two situations shown in Figure 7–23. In cell 1, a transient
signal induces the synthesis of protein A, which is a transcription activator that turns on many genes including its own. In cell 2, a transient signal induces the synthesis of protein R, which is a transcription repressor
that turns of many genes including its own. In which, if either, of these
situations will the descendants of the original cell “remember” that the
progenitor cell had experienced the transient signal? Explain your reasoning.
7–57
Figure 7–24 shows a simple scheme by which three transcription regulators might be used during development to create eight diferent cell
types. How many cell types could you create, using the same rules, with
four diferent transcription regulators? MyoD is a transcription regulator that can trigger the entire program of muscle diferentiation when
expressed in ibroblasts. How might you accomodate this observation
into the scheme shown in Figure 7–24?
Cro gene
NO Cro GENE
TRANSCRIPTION
(B) LYTIC STATE
cI gene
Cro gene
NO cI GENE
TRANSCRIPTION
Figure 7–22 Regulation of bacteriophage
lambda replication by cI and Cro (Problem
7–55). (A) The prophage state. (B) The
lytic state.
DATA HANDLING
7–58
he protein encoded by the Even-skipped (Eve) gene of Drosophila is a
transcription regulator required for proper segmentation in the middle
(A) CELL 1
OFF
transient
signal
A
transcription activator
A
A
A
A
turns on transcription
of activator mRNA
A
activator protein
turns on its own
transcription
(B) CELL 2
OFF
R
transcription repressor
transient
signal
R
R
R
R
turns on transcription
of repressor mRNA
R
repressor protein
turns off its own
transcription
Figure 7–23 Gene regulatory circuits and
cell memory (Problem 7–56). (A) Induction
of synthesis of transcription activator
A by a transient signal. (B) Induction of
synthesis of transcription repressor R by
a transient signal.
MOLECULAR GENETIC MECHANISMS THAT CREATE AND MAINTAIN SPECIALIZED CELL TYPES
Figure 7–24 An illustration of
combinatorial gene control for
development (Problem 7–57). In this
simple, idealized scheme, a “decision”
to make a new transcription regulator
(numbered symbols) is made after
each cell division. In this scheme, the
daughter cell on the right is induced to
make the new transcription regulator.
Each transcription regulator is assumed
to continue to be expressed after it
is induced, thereby allowing different
combinations of regulatory proteins to be
built up. In this example, eight different
cell types have been created.
INDUCE
TRANSCRIPTION
REGULATOR 1
LEFT
1
RIGHT
INDUCE
TRANSCRIPTION
REGULATOR 2
2
151
1
1
2
INDUCE
TRANSCRIPTION
REGULATOR 3
2
3
2
3
A
B
1
1
1
3
2
2
F
G
1
C
D
E
3
H
of the body. It irst appears about 2 hours after fertilization at a uniform
level in all the embryonic nuclei. Not long after that, it forms a pattern
of seven stripes across the embryo. Each stripe is under the control of a
separate module in the Eve promoter, which provides binding sites for
both repressors and activators of Eve transcription. In the case of stripe
2, there are multiple, overlapping binding sites for activators and repressors (Figure 7–25). Two transcription activators, Hunchback (Hb) and
Bicoid (Bcd), and two transcription repressors, Giant (Gt) and Krüppel (Kr), are required to give the normal pattern. he binding sites for
these proteins have been mapped onto the 670-nucleotide segment
shown in Figure 7–25; deletion of this upstream segment abolishes Eve
expression in stripe 2. he patterns of expression of Hunchback, Bicoid,
Giant, and Krüppel in the embryo are shown in Figure 7–26. It seems that
Eve expression in stripe 2 occurs only in the region of the embryo that
expresses both transcription activators, but neither transcription repressor: a simple enough rule.
To check if this rule is correct, you construct a β-galactosidase reporter
gene driven by a 5 kb upstream segment from the Eve promoter. (his
segment also includes the controlling elements for stripes 3 and 7.) In
addition to the normal upstream element, you make three mutant versions in which several of the binding sites in the Eve stripe 2 control segment have been deleted.
Construct 1.
Deletion of all the Krüppel-binding sites
Construct 2.
Deletion of all the Giant-binding sites
Construct 3.
Deletion of two Bicoid-binding sites (indicated by Δ in
Figure 7–25)
Eve stripe 2
control
element
β-galactosidase gene
REPRESSORS
Kr
Kr Gt
Bcd
ACTIVATORS
Gt
Kr
Bcd
Bcd
Δ
Gt Kr
Kr Kr
Hb Bcd
Δ
Hb
Hb
Figure 7–25 Binding sites for transcription
repressors and activators of Eve stripe 2
(Problem 7–58). The Eve stripe 2 control
segment, with additional DNA upstream
of the Eve promoter, was fused to the
β-galactosidase gene, whose product can
be readily visualized in the embryo.
152
Chapter 7: Control of Gene Expression
expression levels
Eve
stripe 2
Hunchback (Hb)
Giant
(Gt)
Krüppel
(Kr)
Bicoid (Bcd)
anterior
pole
posterior
pole
Figure 7–26 Expression of transcription repressors and
activators of Eve stripe 2 in the Drosophila embryo
(Problem 7–58).
You make lies containing these novel genetic constructs integrated
into their chromosomes and determine the patterns of β-galactosidase
expression in their embryos, which are shown in Figure 7–27.
A. Match the mutant embryos to the mutant constructs.
B. You began these experiments to test the simple rule that Eve expression in stripe 2 occurs in the embryo where the two transcription activators are present and the two transcription repressors are absent. Do the
results with the mutant embryos conirm this rule? Explain your answer.
C. Ofer a plausible explanation for why there is no expression of
β-galactosidase at the anterior pole of mutant embryo D in Figure 7–27.
D. In the Eve stripe 2 control segment, the binding sites for the two transcription activators do not overlap, nor do the binding sites for the two
transcription repressors; however, it is often the case that binding sites
for activators overlap binding sites for repressors. What does this overlap
suggest about the mode of genetic control of Eve in stripe 2, and what
might be the consequences for stripe morphology?
7–59
7–60
Hormone receptors for glucocorticoids, upon hormone binding, become
transcription regulators that activate a speciic set of responsive genes.
he DNA- and hormone-binding sites occupy distinct regions of the
C-terminal half of the glucocorticoid receptor. Hormone binding could
generate a functional DNA-binding protein either by altering receptor
conformation to create a DNA-binding domain or by altering the conformation to uncover a preexisting DNA-binding domain.
hese possibilities were investigated by comparing the activities of
a series of C-terminal deletions (Figure 7–28). Segments of the cDNA
that encode N-terminal portions of the receptor were expressed in cells,
and their ability to activate a chloramphenicol acetyl transferase (Cat)
gene—linked to a glucocorticoid-responsive cis-regulatory sequence—
was measured. As shown in Figure 7–28, the full-length receptor (with its
C-terminus at position 0) displayed CAT activity only in the presence of
the glucocorticoid dexamethasone. Most of the mutant receptors failed
to activate CAT expression in the presence or absence of dexamethasone. In contrast, the four mutant receptors lacking 190, 200, 239, and 270
C-terminal amino acids activated CAT expression in the presence and
absence of dexamethasone.
How do these experiments distinguish between the proposed models
for hormone-dependent conversion of the normal receptor to a DNAbinding form? Does hormone binding create a DNA-binding site or does
it uncover a preexisting one?
One of the key transcription regulators produced by the yeast mating-type
locus is a repressor protein known as α2. In haploid cells of the α mating
type, α2 is essential for turning of a set of genes that are speciic for the
stripe numbers
2 3
7
(A) NORMAL
(B) MUTANT
(C) MUTANT
(D) MUTANT
anterior
posterior
Figure 7–27 Embryonic expression of
β-galactosidase from constructs with
normal or mutated Eve stripe 2 control
elements (Problem 7–58).
MOLECULAR GENETIC MECHANISMS THAT CREATE AND MAINTAIN SPECIALIZED CELL TYPES
0
27
1
3
10
12
1
33
28
277
0
23
9
20
0
19
0
18
0
position of C-terminus in mutant receptors
glucocorticoid receptor
N-terminus
DNA
dexamethasone
CAT ASSAY RESULTS
3-acetylchloramphenicol
1-acetylchloramphenicol
chloramphenicol
dexamethasone
– + – + – + – + – + – + – + – + – + – +
331 287 270 239 200 190 180 123 101 27
– +
0
missing C-terminal amino acids
153
Figure 7–28 Effect of C-terminal deletions
on the activity of the glucocorticoid
receptor (Problem 7–59). The schematic
diagram at the top illustrates the
position of the DNA-binding site (DNA)
and the glucocorticoid-binding site
(dexamethasone) in the receptor, as
well as the positions of the C-terminal
deletions. The lower diagram shows
the results of a standard CAT
assay. The lowest spot is unreacted
chloramphenicol; the upper spots
show the attachment of acetyl groups
to one or the other of two positions
on chloramphenicol. The presence (+)
or absence (–) of dexamethasone is
indicated below appropriate lanes.
a mating type. In a/α diploid cells, the α2 repressor collaborates with the
product of the a1 gene to turn of a set of haploid-speciic genes in addition to the a mating-type-speciic genes. Two distinct but related types of
conserved DNA sequences are found upstream of these two sets of regulated genes: one in front of the a-speciic genes and the other in front
of the haploid-speciic genes. Given the relatedness of these upstream
sequences, it is most likely that α2 binds to both; however, its binding
properties must be modiied in some way by the a1 protein before it can
recognize the haploid-speciic sequence. You wish to understand the
nature of this modiication. Does a1 catalyze covalent modiication of α2,
or does it modify α2 by binding to it stoichiometrically?
To study these questions, you perform three types of experiment. In
the irst, you measure the binding of a1 and α2, alone and together, to the
two kinds of upstream regulatory DNA sites. As shown in Figure 7–29,
a1 alone does not bind radiolabeled DNA fragments that contain either
regulatory site (Figure 7–29, lane 2), whereas α2 binds to a-speciic fragments, but not to haploid-speciic fragments (lane 3). he mixture of a1
and α2 binds to a-speciic and haploid-speciic fragments (lane 4).
In the second set of experiments, you add a vast excess of unlabeled
DNA containing the a-speciic sequence, along with the mixture of
a1 and α2 proteins. Under these conditions, the haploid-speciic fragment is still bound (Figure 7–29, lane 5). Similarly, if you add an excess
of unlabeled haploid-speciic DNA, the a-speciic fragment is still bound
(lane 6).
In the third set of experiments, you vary the ratio of a1 relative to
α2. When α2 is in excess, binding to the haploid-speciic fragment is
decreased (Figure 7–29, lane 7); when a1 is in excess, binding to the
a-speciic fragment is decreased (lane 8).
α2
−
−
+
+
+
+
+
+
a1
−
+
−
+
+
+
low high
unlabeled a-specific DNA
−
−
−
−
+
−
−
−
unlabeled haploid-specific DNA
−
−
−
−
−
+
−
−
1
2
3
4
5
6
7
8
haploid-specific
a-specific
Figure 7–29 Binding of transcription
regulators to fragments of DNA
containing the a-speciic or haploidspeciic cis-regulatory sequences
(Problem 7–60). Various combinations
of regulators were incubated with a
mixture of a-speciic and haploid-speciic
radioactive Dna fragments (shown in
lane 1). at the end of the incubation, the
samples were precipitated with antibody
against the proteins, and the Dna
fragments in the precipitate were run on
the gel. The gel was then placed against
x-ray ilm to expose the positions of the
radioactive Dna fragments.
154
Chapter 7: Control of Gene Expression
A. In the presence of a1, is α2 present in two forms with diferent binding
speciicities, or in one form that can bind to both regulatory sequences?
How do your experiments distinguish between these alternatives?
B. An α2 repressor with a small deletion in its DNA-binding domain does
not bind to DNA fragments containing the haploid-speciic sequence.
When this mutant protein is expressed in a diploid cell along with normal α2 and a1 proteins, the haploid-speciic genes are turned on. (hese
genes are normally of in a diploid.) In the light of this result and your
other experiments, do you consider it more likely that a1 catalyzes a covalent modiication of α2, or that a1 modiies α2 by binding to it to form an
a1–α2 complex?
MECHANISMS THAT REINFORCE CELL MEMORY
IN PLANTS AND ANIMALS
TERMS TO LEARN
CG island
DNA methylase
DNA methylation
epigenetic inheritance
genomic imprinting
monoallelic gene expression
X-inactivation
X-inactivation center (XIC)
DEFINITIONS
Match each deinition below with its term from the list above.
7–61
Transcriptional silencing of gene expression on one of the two X chromosomes in the somatic cells of female mammals.
7–62
Addition of a –CH3 group to cytosines in CG sequences in vertebrate
DNA.
7–63
Heritable diference in the phenotype of a cell or an organism that does
not result from changes in the nucleotide sequence of the DNA.
7–64
Expression from only one of the two copies of a gene in the diploid
genome.
7–65
Situation in which one copy of a gene is either expressed or not expressed
in the embryo, depending on which parent it is inherited from.
7–66
Long region of DNA with a much greater than average density of CG
sequences, which usually remain unmethylated.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
7–67
Because CG sequences in one strand are paired with GC sequences in
the other strand, two diferent methyl transferases are required to place
methyl groups on the two strands.
7–68
CG islands are thought to have arisen during evolution because they
were associated with portions of the genome that remained unmethylated in the germ line.
7–69
In most diferentiated tissues, daughter cells retain a memory of gene
expression patterns that were present in the parent cell through mechanisms that do not involve changes in the sequence of their genomic
DNA.
MECHANISMS THAT REINFORCE CELL MEMORY IN PLANTS AND ANIMALS
(A)
(B)
THOUGHT PROBLEMS
7–70
Maintenance methyl transferase, de novo DNA methyl transferases, and
demethylating enzymes play crucial roles in the changes in methylation
patterns during development. Starting with the unfertilized egg, describe
in a general way how these enzymes bring about the observed changes in
genomic DNA methylation.
7–71
Examine the two pedigrees shown in Figure 7–30. One results from
deletion of a maternally imprinted autosomal gene. he other pedigree
results from deletion of a paternally imprinted autosomal gene. In both
pedigrees, afected individuals (red symbols) are heterozygous for the
deletion. hese individuals are afected because one copy of the chromosome carries an imprinted, inactive gene, while the other carries a deletion of the gene. Dotted yellow symbols indicate individuals that carry
the deleted locus, but do not display the mutant phenotype. Which pedigree is based on paternal imprinting and which on maternal imprinting?
Explain your answer.
7–72
Imprinting occurs only in mammals, and why it should exist at all is a
mystery. One idea is that it represents an evolutionary end point in a tug
of war between the sexes. In most mammalian species, a female can mate
with multiple males, generating multiple embryos with diferent fathers.
If one father could cause more rapid growth of his embryo, it would prosper at the expense of the other embryos. While this would be good for the
father’s genes (in an evolutionary sense), it would drain the resources of
the mother, potentially putting her life at risk (not good for her genes).
hus, it is in the mother’s interest to counter these paternal efects with
maternal changes that limit the growth of the embryo.
Based on this scenario, decide whether the Igf 2 gene, whose product—insulin-like growth factor-2—is required for prenatal growth, is
more likely to be imprinted in the male or in the female.
DATA HANDLING
7–73
You are studying the role of DNA methylation in the control of gene
expression, using the human γ-globin gene as a test system (Figure
7–31). Globin mRNA can be detected when this gene is integrated into
the genome of mouse ibroblasts, even though it is expressed at much
lower levels than in red blood cells. If the gene is methylated at all 27 of
its CG sites before integration, its expression is blocked completely. You
are using this system to decide whether a single critical methylation site
is suicient to determine globin expression.
155
Figure 7–30 Pedigrees relecting maternal
and paternal imprinting (Problem 7–71).
In one pedigree, the gene is paternally
imprinted; in the other, it is maternally
imprinted. In generations 3 and 4, only
one of the two parents in the indicated
matings is shown; the other parent
is a normal individual from outside
this pedigree. affected individuals are
represented by red circles for females
and red squares for males. Dotted
yellow symbols indicate individuals that
carry the deletion but do not display the
phenotype.
156
Chapter 7: Control of Gene Expression
exons 1 & 2
exon 3
transcription
unmethylated
methylated
A
0
B
0
C
100
D
0
E
0
F
10
transcript
11
12 13
14 15
CG sequences around
promoter
16
Figure 7–31 Effects of methylation on transcription of the γ-globin gene (Problem
7–73). The methylated segments of the gene are shown in blue; unmethylated
segments are shown in red. The six CG sites around the promoter are shown in
more detail below the constructs. The level of expression of γ-globin RNA from
each construct is shown on the right as a percentage of the expression from a
fully unmethylated construct.
7–74
In female mammals, one X chromosome in each cell is chosen at random
for inactivation early in development. X-inactivation, which involves
more than 1000 genes in humans, is crucial for equalizing expression
of X-chromosome genes in males and females. A critical clue to the
mechanism of X-inactivation came from the isolation of a large number of cDNAs for genes on the human X chromosome. heir expression
patterns were characterized in cells from normal males and females, in
cells from individuals with abnormal numbers of X chromosomes, and in
rodent:human hybrid cell lines that retained either one inactive human
X chromosome (Xi) or one active human X chromosome (Xa). Among all
these cDNAs, there were three patterns of expression, as illustrated by
cDNAs A, B, and C in Figure 7–32.
A. For each pattern of expression, decide whether the gene is expressed
from the active X, the inactive X, or both. Which pattern do you expect to
be the most common? Which pattern is the most surprising?
B. From the results with cells from abnormal individuals, formulate a rule
as to how many chromosomes are inactivated, and how many remain
active during X-inactivation.
a
i
X
XO
XX
Y
XX
XX
XX X
XX
X Y
XY
GENOTYPE
XX
You create several diferent γ-globin constructs that are unmethylated
in speciic regions of the gene. hese constructs are illustrated in Figure
7–31, with the methylated regions shown in blue. he arrangement of six
methylation sites around the promoter is shown below the constructs.
Sites 11, 12, and 13 are unmethylated in construct E; sites 14, 15, and 16
are unmethylated in construct D; and all six sites are unmethylated in
construct F. You integrate these constructs in mouse ibroblasts, grow cell
lines containing individual constructs, verify their methylation patterns,
and measure γ-globin RNA synthesis relative to cell lines containing the
fully unmethylated construct (Figure 7–31).
Does the γ-globin transcription associated with the cell lines containing the constructs (Figure 7–31) indicate that a single critical site of methylation determines whether the gene is expressed? How can you tell?
cDNA
C
A
B
normal
cells
abnormal
cells
hybrid
cells
Figure 7–32 Northern analysis of gene
expression from cells with different
numbers and types of X chromosomes
(Problem 7–74). RNA from cells was run
out on gels, blotted onto nitrocellulose,
probed with a mixture of radioactive
cDNAs A, B, and C, and visualized by
autoradiography. The positions of the
RNA bands that correspond to genes A,
B, and C were determined in a separate
experiment.
POST-TRANSCRIPTIONAL CONTROLS
157
POST-TRANSCRIPTIONAL CONTROLS
TERMS TO LEARN
alternative RNA splicing
internal ribosome entry site (IRES)
post-transcriptional control
regulated nuclear transport
RNA editing
DEFINITIONS
Match each deinition below with its term from the list above.
7–75
Production of a functional RNA by insertion or alteration of individual
nucleotides in an RNA transcript after it has been synthesized.
7–76
A way to generate diferent proteins from the same gene by combining
diferent segments of the initial RNA transcript to make distinct mRNAs.
7–77
General term for a regulatory event that occurs after RNA polymerase has
bound to the gene’s promoter and begun RNA synthesis.
7–78
A sequence in the interior of an mRNA that folds into structures that bind
translation initiation proteins.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
7–79
If the site of transcript cleavage and poly-A addition for a particular RNA
in one cell is downstream of the site used for cleavage and poly-A addition for the same RNA in a second cell, the protein produced from the
longer polyadenylated RNA will necessarily contain additional amino
acids at its C-terminus.
7–80
In one extreme case, a single gene in Drosophila—the Dscam gene—has
the potential to produce more than 38,000 diferent proteins by alternative splicing; thus, the complexity of this one gene rivals the complexity
of the whole human genome.
THOUGHT PROBLEMS
7–81
RNA polymerase II commonly terminates transcription of the HIV (the
human AIDS virus) genome a few hundred nucleotides after it begins,
unless helped along by a virus-encoded protein called Tat, which binds
to a speciic hairpin structure in the nascent viral RNA. Tat then recruits
a collection of proteins, including the protein kinase Cdk9, which phosphorylates RNA polymerase, enhancing its ability to continue transcription. Flavopiridol is the most potent inhibitor of Cdk9 yet discovered; it
blocks Cdk9-mediated phosphorylation. Would you expect lavopiridol
to interfere with HIV transcription? Why or why not?
7–82
Several distinct mechanisms for mRNA localization have been discovered. hey all require speciic sequences in the mRNA itself, usually in
the 3ʹ untranslated region (UTR). Briely outline three mechanisms by
which cellular mRNAs might become localized in the cell.
7–83
Regulation of ferritin translation is controlled by the interaction between
a hairpin structure in the mRNA, termed an iron-response element
(IRE), and the iron-response protein (IRP) that binds to it. When the IRE
is bound by the IRP, translation is inhibited. he location of the IRE in the
mRNA is critical for its function. To work properly, it must be positioned
near the 5ʹ end of the mRNA. If it is moved more than 60 nucleotides
downstream of the cap structure, the IRE no longer inhibits translation.
158
Chapter 7: Control of Gene Expression
Why do you suppose there is such a critical position-dependence for regulation of translation by IRE and IRP?
7–84
Although essential for the cell, iron is also potentially toxic. It is maintained at optimal levels in mammalian cells by the actions of three proteins. Transferrin binds to extracellular iron ions and delivers them to
cells; the transferrin receptor binds iron-loaded transferrin and brings it
into the cell; and ferritin binds iron (up to 4500 atoms in the internal cavity in each complex) to provide an intracellular storage site.
he regulation of the transferrin receptor, like that of ferritin (see
Problem 7–83), is accomplished by IREs and IRPs, and, in both cases,
iron binding to IRPs prevents their binding to IREs. Nevertheless, the
mechanism of transferrin-receptor regulation is quite diferent from that
of ferritin. he transferrin receptor mRNA contains ive IREs that are all
located in the 3ʹ untranslated region of the mRNA (instead of at the 5ʹ end,
as in ferritin mRNA). In addition, binding of IRPs to these IREs increases
the translation of transferrin receptor (as opposed to a decrease for ferritin).
A. Does this opposite regulation of ferritin and transferrin receptor in
response to iron levels make biological sense? Consider the consequences of high and low iron levels.
B. In the presence of iron, the transferrin receptor mRNA is rapidly degraded;
in the absence of iron, it is stable. Can you suggest a mechanism for how
iron levels might be linked to the stability of transferrin receptor mRNA?
7–85
Vg1 mRNA encodes a member of the TGFβ family of growth factors,
which is important for mesoderm induction. In Xenopus, Vg1 mRNA
is localized to the vegetal pole of the oocyte. Translation of Vg1 mRNA
does not occur until a late stage, after localization is complete. Analysis
of the 3ʹ untranslated region identiied two elements: a localization element and a UA-rich translation control element. Translational regulation
was shown to be independent of the poly-A tail. By contrast, translational
repression was abolished when an internal ribosome entry site (IRES)
sequence was inserted into the mRNA upstream of the coding sequence.
How do you suppose translation of Vg1 mRNA is controlled?
DATA HANDLING
7–86
A repeated hexanucleotide element, TGCATG, has been shown to regulate the tissue-speciic splicing of the ibronectin alternative exon EIIIB.
You wonder if it might also regulate alternative splicing in the calcitonin/
CGRP gene (Figure 7–33). In thyroid cells, an mRNA containing exons
1, 2, 3, and 4 encodes calcitonin. In neuronal cells, an mRNA containing
exons 1, 2, 3, 5, and 6 encodes calcitonin gene-related peptide (CGRP).
When you examine the calcitonin/CGRP gene you ind ive copies
of a related repeat, GCATG, within 500 nucleotides of the exon-4 splice
site (Figure 7–33). To analyze their potential role in alternative splicing,
you make several calcitonin/CGRP constructs in which the repeats have
been altered singly or in combination. You transfect the constructs into
HeLa cells, which normally give the calcitonin splicing pattern, and into
F9 cells, which normally give the CGRP splicing pattern. You ind that
no single mutation alters the pattern seen with wild type (not shown);
however, various combinations of the mutations have dramatic efects
(Figure 7–33).
A. Why do you suppose the most dramatic changes were seen in HeLa cells
rather than in F9 cells?
B. Is there a particular combination of GCATG repeats that is critical for
proper calcitonin mRNA production in HeLa cells? If so, what is it?
7–87
In humans, two closely related forms of apolipoprotein B (ApoB) are
found in blood as constituents of the plasma lipoproteins. ApoB48,
POST-TRANSCRIPTIONAL CONTROLS
1 kb
HeLa
cells
159
F9
cells
CGRP
calcitonin
1
2
3
poly A
poly A
4
5
6
HeLa
F9
GCATG repeat 1
2
3
4
5
CALC CGRP
CALC CGRP
wild type
construct A
construct B
construct C
construct D
construct E
construct F
+
−
−
−
−
−
+
+
−
−
−
−
−
+
+
−
−
−
−
+
−
+
−
+
−
+
+
−
++++
−
−
++++
++++
−
−
++++
+
+++
−
++++
++++
−
−
−
−
−
−
−
++
+
−
+
+
−
−
+
++++
++++
++++
++++
++++
++++
++
Figure 7–33 Effects of mutations in
GCATG repeats on alternative splicing
of calcitonin/CGRP pre-mRNA (Problem
7–86). In Hela cells, exon 3 is spliced
to exon 4 to make calcitonin mRNA; in
F9 cells, exon 3 is spliced to exon 5 to
make CGRP mRNA. The positions of
the GCATG repeats around exon 4 are
indicated. In the various constructs, the
presence of a GCATG repeat is indicated
by a plus, and its absence by a minus.
Production of calcitonin- (CAlC) and
CGRP-speciic spliced rna is indicated
relative to wild type as follows: ++++ =
80–100%, +++ = 60–80%, ++ = 40–60%,
+ = 20–40%, and – = 0–20%.
which has a molecular mass of 48 kilodaltons, is synthesized by the intestine and is a key component of chylomicrons, the large lipoprotein particles responsible for delivery of dietary triglycerides to adipose tissue
for storage. ApoB100, which has a molecular mass of 100 kilodaltons, is
synthesized in the liver for formation of the much smaller, very-low-density lipoprotein particles used in the distribution of triglycerides to meet
energy needs. A classic set of studies deined the surprising relationship
between these two proteins.
Sequences of cloned cDNA copies of the mRNAs from these two tissues revealed a single diference: cDNAs from intestinal cells had a T, as
part of a stop codon, at a point where the cDNAs from liver cells had a C,
as part of a glutamine codon (Figure 7–34). To verify the diferences in the
mRNAs and to search for corresponding diferences in the genome, RNA
and DNA were isolated from intestinal and liver cells and then subjected
to PCR ampliication, using oligonucleotides that lanked the region of
interest. he ampliied DNA segments from the four samples were tested
for the presence of the T or C by hybridization to oligonucleotides containing either the liver cDNA sequence (oligo-Q) or the intestinal cDNA
sequence (oligo-STOP). he results are shown in Table 7–2.
Are the two forms of ApoB produced by transcriptional control from
two diferent genes, by a processing control of the RNA transcript from
a single gene, or by diferential cleavage of the protein product from a
single gene? Explain your reasoning.
7–88
he level of β-tubulin gene expression is established in cells by an unusual regulatory pathway, in which the intracellular concentration of free
tubulin dimers (composed of one α-tubulin and one β-tubulin subunit)
regulates the rate of new β-tubulin synthesis. he initial evidence for
such an autoregulatory pathway came from studies with drugs that cause
assembly or disassembly of all cellular tubulin. For example, treatment
of cells with colchicine, which causes microtubule depolymerization into
liver
cDNA
M I Q F D
ATGATACAATTTGAT
ApoB mRNA
TABLE 7–2 Hybridization of specific oligonucleotides to the amplified
segments from liver and intestine RNA and DNA (Problem 7–87).
rna
intestine
cDNA
Dna
Intestine
Liver
Intestine
Liver
Oligo-Q
+
–
+
+
Oligo-STOP
–
+
–
–
Hybridization is indicated by +; absence of hybridization is indicated by –.
ATGATATAATTTGAT
M I *
Figure 7–34 Location of the sequence
difference in cDna clones from apoB
rna isolated from liver and intestine
(Problem 7–87). The encoded amino acid
sequences, in the one-letter code, are
shown aligned with the cDna sequences.
The asterisk indicates a stop codon.
160
Chapter 7: Control of Gene Expression
Figure 7–35 Effects of mutations on the regulation of β-tubulin mRNA
stability (Problem 7–88). The wild-type sequence for the irst 12
nucleotides of the coding portion of the gene is shown at the top, and the
irst four amino acids beginning with methionine (M) are indicated above
the codons. The nucleotide changes in the 12 mutants are shown below;
only the altered nucleotides are indicated. regulation of mrna stability
is shown on the right: + indicates wild-type response to changes in
intracellular tubulin concentration and – indicates no response to changes.
Vertical lines mark the position of the irst nucleotide in each codon.
tubulin dimers, represses β-tubulin synthesis 10-fold. Autoregulation of
β-tubulin synthesis by tubulin dimers is accomplished at the level of
β-tubulin mRNA stability. he irst 12 nucleotides of the coding portion
of the mRNA were found to contain the site responsible for this autoregulatory control.
Since the critical segment of the mRNA involves a coding region, it
is not clear whether the regulation of mRNA stability results from the
interaction of tubulin dimers with the RNA or with the nascent protein.
Either interaction might plausibly trigger a nuclease that would destroy
the mRNA.
hese two possibilities were tested by mutagenizing the regulatory
region on a cloned version of the gene. he mutant genes were then
expressed in cells, and the stability of their mRNAs was assayed in the
presence of excess free tubulin dimers. he results from a dozen mutants
that afect a short region of the mRNA are shown in Figure 7–35.
Does the regulation of β-tubulin mRNA stability result from an interaction with the RNA or from an interaction with the encoded protein?
Explain your reasoning.
7–89
In nematodes, the choice between spermatogenesis and oogenesis in the
hermaphrodite germ line depends on translational regulation of the Tra2
and Fem3 genes, as shown in Figure 7–36. When they are expressed, Tra2
promotes oogenesis and Fem3 promotes spermatogenesis. Translation
of each gene’s mRNA is regulated by the binding of proteins to elements
within their 3ʹ untranslated regions. In each case, the protein-bound
mRNA, although stable, is not eiciently translated and has a short polyA tail. In each case, the mRNA in its unbound form is translationally
active and has a long poly-A tail. How do you suppose that the lengths of
the poly-A tails might afect the eiciency of translation of these mRNAs?
7–90
To investigate the molecular mechanism by which the Tra2 regulatory
elements (TGEs) control translation, you inject Tra2 mRNAs into Xenopus oocytes and follow their fate. You have shown that an oocyte protein
binds to the TGEs and inhibits translation of injected Tra2 mRNA. Now
you wish to determine how mRNAs with bound proteins come to have
short poly-A tails. To investigate this question, you prepare two kinds
(A) Tra2 mRNA
translation OFF
TGE
TGE
short
poly A
translation ON
+
-
-------A---------------C-------
+
-
---------G---------------G---------TA---------
+
-
-----C-C------------G--A-------
+
-
-----G---T----------C----G-----
+
sperm
long poly A
Fem3
translation ON
long poly A
Tra2
-----C---------------C---------
(B) Fem3 mRNA
GLD1 GLD1
Tra2
regulation
M R E I
--ATGAGGGAAATC-+
-----T----------
TGE
TGE
eggs
Fem3
translation OFF
PME
sperm
FBF1
PME
NOS
short
poly A
eggs
Figure 7–36 Translational control of the choice between spermatogenesis and oogenesis (Problem
7–89). (a) Tra2 mrna. (B) Fem3 mrna. Control elements within the 3ʹ untranslated regions of the
two genes are shown, along with the speciic proteins that bind to those elements.
REGULATION OF GENE EXPRESSION BY NONCODING RNAs
length
of poly A
mRNA without TGEs
1
2
4
8
16
32
64
12
8
25
6
51
102
2
1 4
2
4
8
16
32
64
12
8
25
6
51
102
24
mRNA with TGEs
number
of cells
65
161
Figure 7–37 Injection of radiolabeled
mRNAs with and without TGEs into
Xenopus one-cell embryos (Problem
7–90). The lengths of the poly-A tails are
indicated on the left. Cell number refers
to the number of cells in the embryos
when they were harvested.
0
of radiolabeled Tra2 mRNA: one with the normal pair of TGE elements,
and one with those elements deleted (see Figure 7–36A). Each RNA has
a tail of 65 A nucleotides. You inject these mRNAs into one-cell Xenopus
embryos and then re-isolate the mRNAs at various cell stages thereafter.
he re-isolated radiolabeled RNA was analyzed by gel electrophoresis
and autoradiography (Figure 7–37).
A. Do the TGEs alter the overall stability of the mRNAs; that is, are the
mRNAs destroyed more rapidly in the presence or absence of the TGEs?
How can you tell?
B. Do the TGEs inluence the length of the poly-A tails? How can you tell?
7–91
You are skeptical that IRESs really allow direct binding of the eukaryotic
translation machinery to the interior of an mRNA. As a critical test of
this notion, you prepare a set of linear and circular RNA molecules, with
and without IRESs (Figure 7–38A and B). You translate these various
RNAs in rabbit reticulocyte lysates and display the translation products
by sodium dodecyl sulfate (SDS) gel electrophoresis (Figure 7–38C). Do
these results support or refute the idea that IRESs allow ribosomes to initiate translation of mRNAs in a cap-independent fashion? Explain your
answer.
(A) LINEAR mRNA
UAA
AUG
IRES
TRANSLATE
20 kd protein
(B) CIRCULAR mRNA
IRES
AUG
UAA
TRANSLATE
23 kd protein
REGULATION OF GENE EXPRESSION BY
NONCODING RNAs
TERMS TO LEARN
CRISPR
crRNAs
long noncoding RNA (lncRNA)
microRNA (miRNA)
piRNA (piwi-interacting RNA)
RNA interference (RNAi)
small interfering RNA (siRNA)
DEFINITIONS
Match each deinition below with its term from the list above.
7–92
Natural defense mechanism in many organisms that is directed against
foreign RNA molecules, especially those that occur in double-stranded
form.
7–93
A class of short noncoding RNAs that regulate gene expression; roughly
one-third of human genes are thought to be regulated in this way.
7–94
Small RNAs that transcriptionally silence intact transposon genes and
destroy any mRNA produced by them.
7–95
A defense mechanism in bacteria that allows them to destroy viral invaders they have seen before.
7–96
An RNA longer than 200 nucleotides that does not encode a protein.
(C) TRANSLATION ASSAY
linear
circular
IRES
+
−
+
−
+
+
−
−
−
+
−
−
−
+
−
kd
23
20
Figure 7–38 Analysis of effects of IRESs
on translation (Problem 7–91). (A) linear
mRNA that contains an IRES. The
structure of the mRNA without the IRES
was the same. (B) Circular mRNA that
contains an IRES. The structure of the
mRNA without the IRES was the same.
Circular RNAs were prepared by ligating
the ends together; they were puriied
from the linear starting molecules by gel
electrophoresis. (C) Display of translation
products from various species of mrna.
162
Chapter 7: Control of Gene Expression
mouse
chromosome 17
rat
chromosome 20
human
chromosome 6
Figure 7–39 Conservation of piRNA
clusters among mammalian species
(Problem 7–101). The igure is a
schematic depiction of the distribution of
pirnas cloned from a conserved region
of the mouse, rat, and human genomes.
segments that are transcribed from the
top strand of the Dna are shown in blue;
segments transcribed from the bottom
strand are shown in red. The three
genomes are aligned according to this
pattern of transcription.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
7–98
Although the functions of lncRNAs are still mysterious, it is now clear that
they do not function as scafolds for binding groups of proteins.
7–99
piRNAs and crRNAs serve analogous functions; they defend against foreign invaders.
THOUGHT PROBLEMS
7–100
7–101
7–102
List and briely discuss three features that make miRNAs especially useful regulators of gene expression.
piRNAs have several general characteristics: they are roughly 29–30
nucleotides in length, they have a strong bias for U in the irst nucleotide,
they bind to the Piwi class of Argonaute proteins, they are found mostly
in germ cells, and they are clustered in conserved regions of the mouse,
rat, and human genomes (Figure 7–39). hese characteristics suggest
an important function, but the role of not a single piRNA has yet been
deined. Why do you suppose that many biologists are convinced that
piRNAs serve a critical function in organisms?
If you insert a β-galactosidase gene lacking its own transcription control region into a cluster of piRNA genes in Drosophila, you ind that
β-galactosidase expression from a normal copy elsewhere in the genome
is strongly inhibited in the ly’s germ cells. If the inactive β-galactosidase
gene is inserted outside the piRNA gene cluster, the normal gene is properly expressed. What do you suppose is the basis for this observation?
How would you test your hypothesis?
DATA HANDLING
7–103
miRNAs were discovered in nematode worms, but it is still unclear
whether miRNAs reduce protein expression by causing rapid mRNA degradation or by interfering with translation. he Let7 miRNA, for example,
recognizes sites in the 3ʹ end of the Daf12 mRNA, reducing Daf12 protein
synthesis. To investigate how Let7 miRNA controls protein synthesis from
Daf12 mRNA, you make extracts of wild-type larvae at a stage when Let7
miRNA levels are highest, and also make extracts of mutant larvae that do
not make Let7 miRNA at the same stage. Analysis of the polyribosomes
(or polysomes) in normal and mutant larvae revealed no signiicant
diference, nor was there any diference in the distribution of a control
mRNA from the Act1 gene (Figure 7–40A). By contrast, the distribution
of Daf12 mRNA changed signiicantly in the presence of Let7 miRNA
(Figure 7–40B).
(A)
polysome distribution
absorbance 254 nm
Because siRNAs are so widespread among species, they are believed to
be the most ancient form of RNA interference, with miRNAs being a later
reinement.
wild type
Let7
2
(B)
mRNA level (percent of total)
7–97
46
mRNA distribution
20
Daf12
10
0
20
Act1
10
0
2
4
6
8
fraction number
10
12
Figure 7–40 polysome distributions
of mrnas in larval extracts with or
without Let7 mirna (Problem 7–103).
(a) polysome distribution. mrnas with
attached ribosomes (polyribosomes)
were isolated from wild-type and
Let7 mutant larvae and separated by
centrifugation through a sucrose gradient.
The distributions of all mrnas in normal
(wild type; blue) and Let7 mutant larvae
(red) are shown in the top panel. The
monoribosomes are shown to the left of
the dashed line and the polyribosomes
are shown to the right, with a few
individual peaks numbered to indicate
how many ribosomes are present in each
polysome peak. (B) mrna distribution.
The polysome distribution was divided
into 12 fractions, as indicated by the
gray lines. The rna was extracted from
each fraction and the amounts of Daf12
and Act1 mrnas in each fraction were
quantiied. The distributions of Daf12
mrna and Act1 mrna in normal and
Let7 mutant strains are shown below.
REGULATION OF GENE EXPRESSION BY NONCODING RNAs
163
Figure 7–41 Analysis of expression
of X-chromosome-speciic alleles in
undifferentiated and differentiated es
cells (Problem 7–104). analysis of the
a and B alleles from individual cells is
indicated by the brackets.
B A B A B A B A B A B A
mutant ES cells
undifferentiated
B A B A B A B A B A B A
normal ES cells
differentiated
B A B A B A B A B A B A
mutant ES cells
differentiated
A. Does Let7 miRNA cause degradation of Daf12 mRNA? How can you tell?
B. How does Let7 miRNA alter the distribution of Daf12 mRNA? Is Daf12
mRNA in smaller or larger polysomes in the presence of Let7 miRNA?
C. Propose a mechanism by which Let7 miRNA reduces synthesis of Daf12
protein.
7–104
To determine whether the Xist gene, which encodes the Xist lncRNA, is
required for X-inactivation in mice, scientists deleted the Xist gene on
one X chromosome. Using female embryonic stem (ES) cells in which
genes on the two X chromosomes could be distinguished due to polymorphisms, they followed X-inactivation during diferentiation of ES
cells in culture. ES cells normally maintain both X chromosomes in the
active state; however, when they are induced to diferentiate, they randomly inactivate one.
he scientists considered three hypotheses. (1) ES cells mutant for
one Xist gene would fail to register the presence of two X chromosomes
and thus fail to undergo X-inactivation. (2) he Xist knockout would prevent X-inactivation of the targeted X chromosome, thus predisposing the
normal X chromosome to preferential X-inactivation. (3) he mutation
would have no efect on X-inactivation at all.
Using allele-speciic (X-chromosome-speciic) oligonucleotide
probes, they were able to determine which allele of an X-chromosome
gene was expressed in individual cells. As shown in Figure 7–41, they
examined mutant ES cells that were undiferentiated, and mutant and
nonmutant ES cells that had undergone diferentiation. Only a few cells
were examined in Figure 7–41, but analysis of many more conirmed the
patterns shown. he A allele marks the X chromosome whose Xist gene is
intact in the mutant ES cells; the B allele marks the X chromosome from
which the Xist gene was deleted.
Which of the three hypotheses do these results support? Explain your
reasoning.
7–105
Using strand-speciic probes of transcription in the X-inactivation center,
a second gene was found to be transcribed in the opposite direction to
Xist and was named Tsix to indicate its antisense orientation. he Tsix
transcript, like the Xist transcript, has no signiicant open reading frame
and is thought to be a functional lncRNA. As shown in Figure 7–42, the
Tsix transcript extends all the way across the Xist gene. To determine
whether Tsix plays a role in counting X chromosomes, in choosing
which one to inactivate, or in silencing the inactive X, female embryonic
stem (ES) cells were generated in which the promoter for Tsix had been
deleted from one X chromosome. When these ES cells were allowed to
Tsix
promoter
Xist
Tsix
0
10
20
30
40
50 kb
Figure 7–42 arrangement of Xist and
Tsix transcripts in the X-inactivation
center (Problem 7–105). Boxes indicate
the exons of Xist; exons for Tsix are not
shown. The promoter deletion in the Tsix
mutant is indicated.
Chapter 7: Control of Gene Expression
164
diferentiate into females, it was found that the X chromosome with the
Tsix deletion was always inactivated.
A. Is Tsix important for the counting, choice, or silencing of X chromosomes? Explain your answer.
B. Before X-inactivation, Tsix is expressed from both alleles, as is Xist. At the
onset of X-inactivation, Tsix expression becomes conined to the future
active X, whereas Xist expression is restricted to the future inactive X. Can
you suggest some possible ways that Tsix might regulate Xist?
MCAT STYLE
Passage 1 (Questions 7–106 to 7–108)
Embryonic stem cells diferentiate into diverse cell types once they receive the
appropriate signals. Until then they are kept in an undiferentiated state by the
action of three transcription regulators called Oct4, Sox2, and Nanog. An important target of Oct4 and Sox2 is the FGF4 gene. Analysis of the mechanism by which
Oct4 and Sox2 promote transcription of FGF4 provided early clues to how they
work. By analyzing deletions of the DNA regions around the FGF4 gene, scientists identiied a DNA element that was required for the ability of Oct4 and Sox2
to promote transcription of FGF4. he element was located beyond the coding
sequence at the 3ʹ end of the gene. Analysis of the DNA element revealed that it
contained binding sites for Oct4 and Sox2 that were separated by three nucleotides. When the element was mutated to increase the spacing between the binding sites by 3 or more nucleotides, Oct4 and Sox2 could no longer stimulate transcription of FGF4.
he level of expression of Oct4 has strong efects on stem cell diferentiation.
When expressed at normal levels, Oct4 helps maintain the undiferentiated state.
Overexpression of Oct4 by as little as 1.5-fold, however, causes stem cells to differentiate into mesoderm.
What term best describes the DNA element that is required for Oct4 and
Sox2 to promote FGF4 transcription?
A. Enhancer
B. Mediator
C. Promoter
D. TATA box
7–106
Which hypothesis best explains the observation that increasing the spacing between the Oct4 and Sox2 binding sites prevented stimulation of
FGF4 transcription?
A. Oct4 and Sox2 block the spread of repressive chromatin into the 3’ end of
the FGF4 gene.
B. Oct4 and Sox2 must be properly positioned to interact with the RNA polymerase II complex.
C. Precise positioning of the binding sites promotes cooperative binding of
Oct4 and Sox2.
D. he position of Oct4 and Sox2 inluences the spacing of nucleosomes
over the FGF4 gene.
7–107
Which one of the following actions best explains how a minor change in
Oct4 levels could lead to a big change in the diferentiation state of cells?
A. Chromatin modiication
B. Cooperative binding
C. Dimerization
D. Negative feedback
7–108
Passage 2 (Questions 7–109 to 7–111)
Treatment of mouse ibroblast cells with 5-azacytidine causes some of the cells
to diferentiate into muscle cells. A search for proteins that are responsible for
MCAT STYLE
this diferentiation identiied a transcription regulator called MyoD. Transfection
of cells with a vector that expresses MyoD from an inducible promoter caused
eicient diferentiation of ibroblast cells into muscle cells when expression of
MyoD was turned on. Moreover, when expression of MyoD from the vector was
later turned of, the cells remained diferentiated as muscle cells. MyoD controls
the transcription of numerous genes. Some of these genes are turned on early in
the diferentiation process, whereas others are turned on late. Interestingly, MyoD
binds to the promoters of the late genes immediately after MyoD expression is
irst turned on, yet expression of these genes does not occur until later.
Which of the following possible efects of 5-azacytidine would best
account for its ability to cause ibroblasts to diferentiate into muscle
cells?
A. Activation of Mediator complex
B. Inhibition of a DNA demethylase
C. Inhibition of a speciic riboswitch
D. Inhibition of histone modiication
165
(A)
MyoD
early genes
late genes
early genes
late genes
early genes
late genes
early genes
late genes
(B)
MyoD
7–109
Which one of the following mechanisms best explains how cells remain
diferentiated as muscle cells, even after the original pulse of MyoD
expression has been turned of?
A. Cooperative binding
B. Negative feedback
C. Positive feedback
D. Synergistic transcriptional activation
7–110
7–111
Which one of the regulatory circuits in Figure 7–43 best explains all of
the observations regarding the role of MyoD in induction of muscle cell
diferentiation?
(C)
MyoD
(D)
MyoD
Figure 7–43 Four potential MyoD
regulatory circuits (Problem 7–111).
A 1.5 mL Eppendorf Tube.
An Eppendorf tube containing
10 microliters of solution. Most molecular
biology reactions are carried out in these
tiny plastic tubes, which were invented
in 1961 at the Eppendorf hospital
laboratories in the suburbs of Hamburg,
Germany, as part of a system for handling
very small volumes of clinical samples.
These tubes are inert, robust, and
inexpensive.
Chapter 8
167
Analyzing Cells, Molecules,
and Systems
ISOLATING CELLS AND GROWING THEM IN CULTURE
TERMS TO LEARN
hybridoma
monoclonal antibody
CHAPTER
8
IN THIS CHAPTER
ISOLATING CELLS AND
GROWING THEM IN CULTURE
DEFINITIONS
PURIFYING PROTEINS
Match each deinition below with its term from the list above.
ANALYZING PROTEINS
8–1
Antibody secreted by a hybridoma cell line.
8–2
Cell line used in the production of monoclonal antibodies; obtained by
fusing antibody-secreting B cells with cells of a lymphocyte tumor.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
8–3
Laser-capture microdissection permits isolation of individual cells from
a sample of tissue.
8–4
Because a monoclonal antibody recognizes a speciic antigenic site
(epitope), it binds only to the speciic protein against which it was made.
THOUGHT PROBLEMS
8–5
A common step in the isolation of cells from a sample of animal tissue is
to treat it with trypsin, collagenase, and EDTA. Why is such a treatment
necessary, and what does each component accomplish? Why doesn’t
this treatment kill the cells?
8–6
Isolation of cells from tissues, luorescence-activated cell sorting, and
laser-capture microdissection are just a few of the ways for generating
homogeneous cell populations. Why do you suppose it is important to
have a homogeneous cell population for many experiments?
8–7
Distinguish among the terms “primary culture,” “secondary culture,” and
“cell line.”
8–8
Consider the following two statements. “he most important advantage
of the hybridoma technique is that monoclonal antibodies can be made
against molecules that constitute only a minor component of a complex
mixture.” “he most important advantage of the hybridoma technique
is that antibodies that may be present as only minor components in
conventional antiserum can be produced in quantity in pure form as
monoclonal antibodies.” Are these two statements equivalent? Why or
why not?
ANALYZING AND MANIPULATING
DNA
STUDYING GENE EXPRESSION
AND FUNCTION
MATHEMATICAL ANALYSIS
OF CELL FUNCTIONS
168
Chapter 8: Analyzing Cells, Molecules, and Systems
8–9
Do you suppose it would be possible to raise an antibody against another
antibody? Explain your answer.
1
2
3
4
5
6
7
8
9 10
CALCULATIONS
8–10
You want to isolate rare cells that are present in a population at a frequency of 1 in 105 cells, and you need 10 of those cells to do an experiment. If your luorescence-activated cell sorter can sort cells at the rate
of 1000 per second, how long would it take to collect enough rare cells for
your experiment?
Panels of human–rodent cell hybrids that retain one or a few human
chromosomes, parts of human chromosomes, or radiation-induced fragments of human chromosomes have proven enormously useful in mapping genes to deined locations. Now that the human genome has been
sequenced, it is a trivial matter to know a gene’s location if you have a
bit of sequence from the gene. Nevertheless, there are many instances
in which such panels of cells are still invaluable; for example, when you
know a phenotype, but not the identity of the gene responsible for it, as
in the following case.
You wish to map the location of the receptor for feline leukemia virus
type C (FeLV-C), which infects human cells but not rodent cells. Using
a panel of hybrid cells carrying whole chromosomes, you have shown
that FeLV-C infects only those hybrids carrying human chromosome 1.
Using a second panel carrying portions of chromosome 1, you show that
FeLV-C infects several of the hybrid cell lines. he segments of chromosome 1 that are present in the infectable hybrid cell lines are shown in
Figure 8–1. Where on chromosome 1 is the gene for the FeLV-C receptor
located?
PURIFYING PROTEINS
TERMS TO LEARN
column chromatography
fusion protein
high-performance liquid
chromatography (HPLC)
purified cell-free system
DEFINITIONS
Match each deinition below with its term from the list above.
8–12
General term for puriication technique in which a mixture of proteins is
passed through a cylinder containing a porous solid matrix.
8–13
Type of chromatography that uses columns packed with special chromatography resins composed of tiny spheres that attain a high degree of
resolution, even at very fast low rates.
8–14
Artiicial product generated by linking the coding sequences for two different proteins, or protein segments, and expressing the hybrid gene in
cells.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
8–15
p2
p1
DATA HANDLING
8–11
p3
It is possible to pellet hemoglobin by centrifugation at suiciently high
speed.
q1
q2
q3
q4
Figure 8–1 Mapping the gene for the
FelV-C receptor using human–rodent
hybrid cell lines that carry portions of
human chromosome 1 (Problem 8–11).
The hybrid cell lines that could be
infected by FelV-C are shown, with
the portions of human chromosome 1
retained in the individual hybrid cell
lines indicated as blue lines. The
p and q designations refer to a standard
convention (the Paris nomenclature) for
describing chromosome positions. Shortarm locations are labeled p (petit) and
long arms q (queue). Each chromosome
arm is divided into regions labeled
p1, p2, p3, q1, q2, q3, etc., counting
outward from the centromere. Regions
are delimited by speciic landmarks,
which are distinct morphological
features, including the centromere and
certain prominent bands. Regions are
divided into bands labeled p11 (oneone, not eleven), p12, etc. Sub-bands
are designated p11.1, p11.2, etc., and
sub-sub bands are designated p11.11,
p11.12, etc. In all cases, the numbers
increase from the centromere toward
the telomere.
PURIFYING PROTEINS
169
Figure 8–2 Backbone models of
tropomyosin and hemoglobin
(Problem 8–19).
hemoglobin
8–16
tropomyosin
If the beads used in gel-iltration chromatography had pores of a uniform size, proteins would either be excluded from the pores or included
in them, but would not be further fractionated.
THOUGHT PROBLEMS
8–17
Describe how you would use preparative centrifugation to purify mitochondria from a cell homogenate.
8–18
Distinguish between velocity sedimentation and equilibrium sedimentation. For what general purpose is each technique used? Which do you
suppose might be best suited for separating two proteins of diferent
size?
8–19
Tropomyosin, at 93 kd, sediments at 2.6S, whereas the 65-kd protein,
hemoglobin, sediments at 4.3S. (he sedimentation coeicient S is a linear measure of the rate of sedimentation: both increase or decrease in
parallel.) hese proteins are drawn to scale in Figure 8–2. How is it that
the bigger protein sediments more slowly than the smaller one? Can you
think of an analogy from everyday experience that might help you with
this problem?
8–20
Distinguish among ion-exchange chromatography, hydrophobic chromatography, gel-iltration chromatography, and ainity chromatography
in terms of the column material and the basis for separation of a mixture
of proteins.
CALCULATIONS
8–21
he puriication of a protein usually requires multiple steps and often
involves several types of column chromatography. A key component of
any puriication is an assay for the desired protein. he assay can be a
band on a gel, a structure in the electron microscope, the ability to bind
to another molecule, or enzyme activity. he puriication of an enzyme
is particularly instructive because the assay allows one to quantify the
extent of puriication at each step. Consider the puriication of the
enzyme shown in Table 8–1. he total volume, total protein, and total
enzyme activity are shown at each step.
A. For each step in the puriication procedure, calculate the speciic activity
of the enzyme (units of activity per mg of protein). How can you tell that
puriication has occurred at each step?
B. Which of the puriication steps was most efective? Which was least efective?
C. If you were to carry the puriication through additional steps, how would
the speciic activity change? How could you tell from speciic activity
measurements that the enzyme was pure? How might you check on that
conclusion?
170
Chapter 8: Analyzing Cells, Molecules, and Systems
TABLE 8–1 Purification of an enzyme (Problem 8–21).
Procedure
Total volume (ml)
Total protein (mg)
Total activity (units)
1. Crude extract
2000
15,000
150,000
2. Ammonium sulfate precipitation
320
4000
140,000
3. Ion-exchange chromatography
100
550
125,000
4. Gel-filtration chromatography
85
120
105,000
5. Affinity chromatography
8
5
75,000
Speciic activity
(units/mg)
D. If the enzyme is pure at the end of the puriication scheme in Table 8–1,
what proportion of the protein in the starting cell does it represent?
DATA HANDLING
8–22
8–23
In the classic paper that demonstrated the semiconservative replication
of DNA, Meselson and Stahl began by showing that DNA itself will form
a band when subjected to equilibrium sedimentation. hey mixed randomly fragmented E. coli DNA with a solution of CsCl so that the inal
solution had a density of 1.71 g/mL. As shown in Figure 8–3, with increasing length of centrifugation at 70,000 times gravity, the DNA, which was
initially dispersed throughout the centrifuge tube, became concentrated
over time into a discrete band in the middle.
A. Describe what is happening with time and explain why the DNA forms a
discrete band.
B. What is the buoyant density of the DNA? (he density of the solution at
which DNA “loats” at equilibrium deines the “buoyant density” of the
DNA.)
C. Even if the DNA were centrifuged for twice as long—or even longer—the
width of the band remains about what is shown at the bottom of Figure
8–3. Why doesn’t the band become even more compressed? Suggest
some possible reasons to explain the thickness of the DNA band at equilibrium.
he result of gel-iltration chromatography of six roughly spherical proteins is shown in Figure 8–4. he identities of the proteins, their molecular masses, and their elution volumes are indicated in Table 8–2. (he
elution volume identiies when each protein came of the column.)
hours
centrifugal field
0
2.1
4.3
6.4
8.5
10.7
12.8
14.9
17.1
TABLE 8–2 Proteins separated by gel-filtration chromatography
(Problem 8–23).
Protein
Molecular
mass (kd)
Molecular
mass (log)
Elution volume
(ml)
Ribonuclease A
13
4.11
250
Chymotrypsinogen
25
4.40
228
Ovalbumin
43
4.63
199
Bovine serum albumin
67
4.83
176
Aldolase
158
5.20
146
Catalase
232
5.37
123
19.2
21.3
23.5
36.5
43.5
Figure 8–3 Ultraviolet (UV) absorption
photographs showing successive
stages in the banding of E. coli DNA
(Problem 8–22). DNA, which absorbs
UV light, shows up as dark regions in
the photographs. The bottom of the
centrifuge tube is on the right.
PURIFYING PROTEINS
171
absorbance at 280 nm
0.4
Figure 8–4 Elution proile for
proteins fractionated by gel-iltration
chromatography (Problem 8–23). The
absorbance at 280 nm is a measure of
protein concentration. Each of the peaks
is identiied by its elution volume.
199 228
146
0.3
176
250
123
0.2
0.1
0
0
50
100
150
200
250
300
elution volume (mL)
A. Why do the smaller proteins come of the column later than the larger
proteins?
B. Plot molecular mass versus elution volume. Now plot the log of the
molecular mass versus the elution volume. Which plot gives a straight
line? What do you suppose is the basis for that result?
8–24
In preliminary studies, you’ve determined that your partially puriied
protein is stable (retains activity) between pH 5.0 and pH 7.5. On either
side of that pH range, the protein is no longer active. Your advisor now
wants you to do a quick experiment to determine conditions for ionexchange chromatography. He has left instructions for you. First, you’re
supposed to mix a bit of the crude preparation with a small amount of the
ion-exchange resin DEAE-Sepharose™ in a series of bufer solutions that
have a pH between 5.0 and 7.5. Next, you are to pellet the resin and assay
the supernatant for the presence of your protein. Finally, he tells you to
use this information to pick the proper pH to do the ion-exchange chromatography. You have completed the irst two steps and have obtained
the results shown in Figure 8–5. But you are a little uncertain as to how to
use the information to pick the pH for the chromatography.
A. At which end of the pH range is the charge on your protein more positive
and at which end is it more negative? [Over this pH range, the positively
charged amine groups on the DEAE-Sepharose beads (Figure 8–5A) are
unafected.]
B. For the chromatography, should you pick a pH at which the protein binds
to the beads (pH 6.5 to 7.5) or a pH where it does not bind (pH 5.0 to 6.0)?
Explain your choice.
C. Should you pick a pH close to the boundary (that is, pH 6.0 or 6.5) or far
away from the boundary (that is, pH 5.0 or pH 7.5)? Explain your reasoning.
D. How will you carry out ion-exchange chromatography of your protein?
What are the various steps you will use to accomplish the separation of
your protein from others via ion-exchange chromatography?
(A) STRUCTURE OF DEAE-SEPHAROSE
(B) TEST FOR CONDITIONS
protein in supernatant after mixing
+
+
+
−
−
−
C2H5
O CH2 CH2 N
H
protein
solution
C2H5
DEAESepharose
pH 5.0
5.5
6.0
6.5
7.0
7.5
Figure 8–5 Preliminary test to
determine conditions for ion-exchange
chromatography (Problem 8–24).
(A) Structure of the charged amine
groups attached to Sepharose beads.
(B) Results of mixing your protein with
DEAE-Sepharose beads. Samples of the
protein were mixed with DEAE-Sepharose
beads in buffers at a range of pH values,
and then the mixtures were centrifuged
to pellet the beads. The presence of the
protein in the supernatant is indicated by
a +; its absence is indicated by a –.
172
Chapter 8: Analyzing Cells, Molecules, and Systems
ANALYZING PROTEINS
TERMS TO LEARN
chemical biology
nuclear magnetic resonance
(NMR) spectroscopy
SDS polyacrylamide-gel
electrophoresis (SDS-PAGE)
two-dimensional gel electrophoresis
Western blotting (immunoblotting)
x-ray crystallography
DEFINITIONS
Match each deinition below with its term from the list above.
8–25
Analysis of the release of electromagnetic radiation by atomic nuclei in a
magnetic ield, due to lipping of the orientation of their magnetic dipole
moments.
8–26
Technique for protein separation in which the protein mixture is run irst
in one direction and then in a direction at right angles to the irst.
8–27
Technique in which a protein mixture is separated by running it through
a gel containing a detergent that binds to and unfolds the proteins.
8–28
he main technique that has been used to discover the three-dimensional structure of molecules, including proteins, at atomic resolution.
8–29
Technique by which proteins are separated by electrophoresis, immobilized on a paper sheet, and then analyzed, usually by means of a labeled
antibody.
TRUE/FALSE
Decide whether the statement is true or false, and then explain why.
8–30
Given the inexorable march of technology, it seems inevitable that the
sensitivity of detection of molecules will ultimately be pushed beyond
the yoctomole level (10–24 mole).
THOUGHT PROBLEMS
8–31
How is it that smaller molecules move through a gel-iltration column
more slowly than larger molecules, whereas in SDS polyacrylamide-gel
electrophoresis (SDS-PAGE) the opposite is true: larger molecules move
more slowly than small molecules?
8–32
You are set to run your irst SDS polyacrylamide-gel electrophoresis. You
have boiled your samples of protein in SDS in the presence of mercaptoethanol and loaded them into the wells of a polyacrylamide gel. You
are now ready to attach the electrodes. Uh oh, does the positive electrode
(the anode) go at the top of the gel, where you loaded your proteins, or at
the bottom of the gel?
8–33
You hate the smell of mercaptoethanol. Since disulide bonds in intracellular proteins are very rare (see Problem 3–37), you have convinced
yourself that it is not necessary to treat a cytoplasmic homogenate with
mercaptoethanol prior to SDS-PAGE. You heat a sample of your homogenate in SDS and subject it to electrophoresis. Much to your surprise, your
gel looks horrible; it is an ugly smear! You show your result to a fellow
student with a background in chemistry, and she suggests that you treat
your sample with N-ethylmaleimide (NEM), which reacts with free sulfhydryls. You run another sample of your homogenate after treating it
with NEM and SDS. Now the gel looks perfect!
ANALYZING PROTEINS
173
If the vast majority of intracellular proteins don’t have disulide
bonds—and they don’t—why didn’t your original scheme work? And
how does treatment with NEM correct the problem?
8–34
For separation of proteins by two-dimensional polyacrylamide-gel electrophoresis, what are the two types of electrophoresis that are used in
each dimension? Do you suppose it makes any diference which electrophoretic method is applied irst? Why or why not?
8–35
Discuss the following statement: “With the ever-expanding databases
of protein sequences and structures, it will soon be possible to input an
amino acid sequence of an unknown protein and, by analogy to known
proteins, determine its structure and function. hus, it will not be long
before biochemists are put out of work.”
8–36
Hybridoma technology allows one to generate monoclonal antibodies
to virtually any protein. Why is it, then, that genetically tagging proteins
with epitopes is such a commonly used technique, especially since an
epitope tag has the potential to interfere with the function of the protein?
8–37
Speciic activity refers to the amount of radioactivity per unit amount
of substance, most commonly in biology expressed on a molar basis,
for example, as Ci/mmol. [One curie (Ci), which is the standard unit of
radioactive decay, corresponds to 2.22 × 1012 disintegrations per minute
(dpm).] If you examine Table 8–3, you will see that there seems to be an
inverse relationship between maximum speciic activity and half-life. Do
you suppose this is just a coincidence or is there an underlying reason?
Explain your answer.
8–38
You just developed an autoradiograph after a two-week exposure. You
had incubated your protein with a cell-cycle kinase in the presence of
32P-ATP in hopes of demonstrating that it was indeed a substrate for the
kinase. You see the hint of a band on the gel at the right position, but it is
just too faint to be convincing. You show your result to your advisor and
tell him that you’ve put the blot against a fresh sheet of ilm, which you
plan to expose for a longer period of time. He gives you a sideways look
and tells you to do the experiment over again and use more radioactivity.
What’s wrong with your plan to reexpose the blot for a longer time?
CALCULATIONS
8–39
How many copies of a protein need to be present in a cell in order for it
to be visible as a band on an SDS gel? Assume that you can load 100 μg
of cell extract onto a gel and that you can detect 10 ng in a single band
by silver staining. he concentration of protein in cells is about 200 mg/
mL, and a typical mammalian cell has a volume of about 1000 μm3 and
a typical bacterium a volume of about 1 μm3. Given these parameters,
TABLE 8–3 Radioactive isotopes and some of their properties (Problem
8–37).
Radioactive
isotope
Emission
Half-life
Maximum speciic
activity (Ci/mmol)
14C
β particle
5730 years
0.062
3H
β particle
12.3 years
29
35S
β particle
87.4 days
1490
32P
β particle
14.3 days
9120
174
Chapter 8: Analyzing Cells, Molecules, and Systems
calculate the number of copies of a 120-kd protein that would need to be
present in a mammalian cell and in a bacterium in order to give a detectable band on a gel. You might try an order-of-magnitude guess before
you make the calculations.
8–40
How many molecules of your labeled protein are required for detection
by autoradiography? You added 1 μL containing 10 μCi of γ-32P-ATP (a
negligible amount of ATP) to 9 μL of cell extract that had an ATP concentration of 1 mM. You incubated the mixture to allow transfer of phosphate to proteins in the extract. You then subjected 1 μL of the mixture to
SDS-PAGE, dried the gel, and placed it against a sheet of x-ray ilm. After
an overnight exposure you saw a barely detectable band in the location of
your protein. You know from previous experience that a protein labeled
at 1 count per minute per band (1 cpm equals 1 disintegration per minute, dpm, for 32P) will form such a band after an overnight exposure. How
many molecules of labeled protein are in the band, if you assume 1 phosphate per molecule? (Some useful conversion factors for radioactivity are
shown in Table 3 on page 964.)
8–41
You want to know the sensitivity for detection of immunoblotting (Western blotting), using an enzyme-linked second antibody to detect the
antibody directed against your protein (Figure 8–6A). You are using the
mouse monoclonal antibody 4G10, which is speciic for phosphotyrosine residues, to detect phosphorylated proteins. You irst phosphorylate
the myelin basic protein in vitro using a tyrosine protein kinase that adds
one phosphate per molecule. You then prepare a dilution series of the
phosphorylated protein and subject the samples to SDS-PAGE. Next, the
protein is transferred (blotted) onto a nitrocellulose ilter, incubated with
the 4G10 antibody, and washed to remove unbound antibody. he blot
is then incubated with a second goat anti-mouse antibody that carries
horseradish peroxidase (HRP) conjugated to it, and any excess unbound
antibody is again washed away. You place the blot in a thin plastic bag,
add reagents that chemiluminesce when they react with HRP (Figure
8–6A), and place the bag against a sheet of x-ray ilm. When the ilm is
developed, you see the picture shown in Figure 8–6B.
A. Given the amounts of phosphorylated myelin basic protein indicated in
each lane in Figure 8–6B, calculate the detection limit of this method in
terms of molecules of protein per band.
B. Assuming that you were using monoclonal antibodies to detect proteins,
would you expect that the detection limit would depend on the molecular mass of the protein? Why or why not?
(B) IMMUNOBLOT
(A) SCHEMATIC DIAGRAM
myelin basic protein (fmol)
40
light
NH2
O
NH2
O
O−
NH
NH
luminol
O
O−
H2O2
O
HRP
second antibody
MBP
first antibody
nitrocellulose membrane
20
10
5
2.5 1.2 0.6
Figure 8–6 Sensitivity of detection
of immunoblotting (Problem 8–41).
(A) Schematic diagram of the experiment.
MBP stands for myelin basic protein.
In the presence of hydrogen peroxide,
horseradish peroxidase (HRP) converts
luminol to a chemiluminescent molecule
that emits light, which is detected by
exposure of an x-ray ilm. (B) Exposed
ilm of an immunoblot. The number of
femtomoles of myelin basic protein in
each band is indicated.
ANALYZING PROTEINS
175
DATA HANDLING
You have isolated the proteins from two adjacent spots after two-dimensional polyacrylamide-gel electrophoresis and digested them with
trypsin. When the masses of the peptides were measured by MALDI-TOF
mass spectrometry, the peptides from the two proteins were found to
be identical except for one (Figure 8–8). For this peptide, the mass-tocharge (m/z) values difered by 80, a value that does not correspond to a
diference in amino acid sequence. (For example, glutamic acid instead
of valine at one position would give an m/z diference of around 30.) Can
you suggest a possible diference between the two peptides that might
account for the observed m/z diference?
8–44
You have raised four diferent monoclonal antibodies to Xenopus Orc1,
which is a component of the DNA replication origin recognition complex (ORC) found in eukaryotes. You want to use the antibodies to immunopurify other members of ORC. To decide which of your monoclonal
antibodies—TK1, TK15, TK37, or TK47—is best suited for this purpose,
you covalently attach them to beads, incubate them with a Xenopus egg
extract, spin the beads down and wash them carefully, and then solubilize the bound proteins with SDS. You use SDS-PAGE to separate the
solubilized proteins and stain them, as shown in Figure 8–9.
A. From these results, which bands do you think arise from proteins that are
present in ORC?
B. Why do you suppose the various monoclonal antibodies give such diferent results?
C. Which antibody do you think is the best one to use in future studies of
this kind? Why?
D. How might you determine which band on this gel is Orc1?
abundance
abundance
8–43
3706
116
97
67
56
cyclin B
40
ribonucleotide
reductase
35
Figure 8–7 Autoradiograph of
radiolabeled proteins separated by
SDS-PAGE (Problem 8–42). A set of
radiolabeled marker proteins with known
molecular masses is shown in the lefthand lane, along with their molecular
masses in kilodaltons. Radiolabeled
proteins from a sea urchin egg extract
are shown in the right-hand lane. Arrows
mark the bands that correspond to cyclin
B and the small subunit of ribonucleotide
reductase.
A
b
TK 423
1
TK
15
TK
37
TK
47
Figure 8–7 shows an autoradiograph of an SDS-PAGE separation of
radiolabeled proteins in a cell-free extract of sea urchin eggs. Alongside
are shown a set of radiolabeled marker proteins of deined molecular
mass. Two bands that contain known proteins—the small subunit of ribonucleotide reductase and cyclin B—are indicated.
A. Do the standard set of proteins migrate at a rate that is inversely proportional to their molecular masses? hat is to say, would you expect a protein of 35 kd, for example, to migrate twice as far down the gel as a protein
of 70 kd? Do you suppose a plot of log molecular mass versus migration
would give a more linear relationship?
B. How would you use the standard set of proteins to estimate the molecular masses of ribonucleotide reductase and cyclin B? What would you
estimate the molecular masses of these two proteins to be?
C. he sequences of the genes for these two proteins give molecular masses
of 44 kd for ribonucleotide reductase and 46 kd for cyclin B. Can you ofer
some possible reasons why the SDS-PAGE estimate for the molecular
mass of cyclin B is so far of?
m
8–42
kd
kd
116
97
66
45
29
20
18
3786
m/z (mass-to-charge ratio)
Figure 8–8 Masses of peptides measured by MAlDI-ToF
mass spectrometry (Problem 8–43). only the numbered
peaks differ between the two protein samples.
Figure 8–9 Immunoafinity puriication
of Xenopus oRC (Problem 8–44). The
monoclonal antibody mAb423 is speciic
for an antigen not found in Xenopus
extracts and thus serves as a control. The
positions of marker proteins are shown
at the left with their masses indicated in
kilodaltons.
176
Chapter 8: Analyzing Cells, Molecules, and Systems
VP16 activation domain
LexA DNA-binding
domain
VP16
transcription
LacZ gene
LexA-binding site
8–45
Figure 8–10 Activation of transcription by
a hybrid transcription regulator (Problem
8–45).
LacZ gene
he yeast two-hybrid system relies on the cell’s own mechanisms to reveal
protein–protein interactions. his method takes advantage of the modular nature of many transcription regulators, which have one domain that
binds to DNA and another domain that activates transcription. Domains
can be interchanged by recombinant DNA methods, allowing hybrid
transcription regulators to be constructed. hus, the DNA-binding
domain of the E. coli LexA repressor can be combined with the powerful VP16 activation domain from herpesvirus to activate transcription of
genes downstream of a LexA DNA binding site (Figure 8–10).
If the two domains of the transcription regulator can be brought into
proximity by protein–protein interactions, they will activate transcription. his is the key feature of the two-hybrid system. hus, if one member of an interacting pair of proteins is fused to the DNA-binding domain
of LexA (to form the “bait”) and the other is fused to the VP16 activation domain (to form the “prey”), transcription will be activated when
the two hybrid proteins interact inside a yeast cell. It is possible to design
powerful screens for protein–protein interactions, if the gene whose transcription is turned on is essential for growth or can give rise to a colored
product.
To check out the ability of the system to ind proteins with which Ras
interacts, hybrid genes were constructed that contained the LexA DNAbinding domain, one fused to Ras (LexA–Ras) and the other fused to
nuclear lamin (LexA–lamin). A second pair of constructs contained the
VP16 activation domain alone (VP16) or fused to the adenylyl cyclase
gene (VP16–CYR). Adenylyl cyclase is known to interact with Ras and
serves as a positive control; nuclear lamins do not interact with Ras and
serve as a negative control. hese plasmid constructs were introduced
into a strain of yeast containing copies of the His3 gene and the LacZ
gene, both with LexA-binding sites positioned immediately upstream.
Individual transformed colonies were tested for the ability to grow on
a plate lacking histidine, which requires expression of the His3 gene. In
addition, they were tested for ability to form blue colonies (as compared
to the normal white colonies) when grown in the presence of an appropriate substrate (XGAL) for β-galactosidase. he set-up for the experiment is outlined in Table 8–4.
TABLE 8–4 Experiments to test the two-hybrid system (Problem 8–45).
Plasmid constructs
Bait
Prey
lexA–Ras
LexA–lamin
VP16
VP16–CYR
LexA–Ras
VP16
LexA–Ras
VP16–CYR
LexA–lamin
VP16
LexA–lamin
VP16–CYR
Growth on plates lacking histidine
Color on plates with XGAl
ANALYZING AND MANIPULATING DNA
177
A. Fill in Table 8–4 with your expectations. Use a plus sign to indicate growth
on plates lacking histidine and a minus sign to indicate no growth. Write
“blue” or “white” to indicate the color of colonies grown in the presence
of XGAL.
B. For any combinations of bait and prey in the table that you expect to
confer growth in the absence of histidine and to form blue colonies with
XGAL, sketch the structure of the active transcription regulator on the
LacZ gene.
C. If you want two proteins to be expressed in a single polypeptide chain,
what must you be careful to do when you fuse the two genes together?
ANALYZING AND MANIPULATING DNA
TERMS TO LEARN
bacterial artificial chromosome (BAC)
cDNA clone
cDNA library
deep RNA sequencing (RNA-seq)
dideoxy sequencing (Sanger
sequencing)
DNA cloning
DNA library
genome annotation
genomic library
hybridization
open reading frame (ORF)
plasmid vector
polymerase chain reaction (PCR)
recombinant DNA technology
restriction nuclease
DEFINITIONS
Match each deinition below with its term from the list above.
8–46
Small, circular DNA molecule that replicates independently of the
genome and can be used for DNA cloning.
8–47
A collection of clones that contain a variety of DNA segments from the
genome of an organism.
8–48
he process of marking out all the genes in a genome and ascribing a
biological function to each.
8–49
One of a large number of enzymes that can cleave a DNA molecule at any
site where a speciic short sequence of nucleotides occurs.
8–50
A DNA clone of a DNA copy of an mRNA molecule.
8–51
Technique for generating multiple copies of speciic regions of DNA by
the use of sequence-speciic primers and multiple cycles of DNA synthesis.
8–52
he sequencing of the entire repertoire of RNA from a cell or tissue.
8–53
Prokaryotic cloning vector that can accommodate large pieces of DNA
up to 1 million base pairs.
8–54
he process whereby two complementary nucleic acid strands form a
double helix.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
8–55
Bacteria that make a speciic restriction nuclease for defense against
viruses have evolved in such a way that their own genome does not contain the recognition sequence for that nuclease.
178
Chapter 8: Analyzing Cells, Molecules, and Systems
8–56
Pulsed-ield gel electrophoresis uses a strong electric ield to separate
very long DNA molecules, stretching them out so that they travel endirst through the gel at a rate that depends on their length.
8–57
By far the most important advantage of cDNA clones over genomic clones
is that they can contain the complete coding sequence of a gene.
8–58
If each cycle of PCR doubles the amount of DNA synthesized in the previous cycle, then 10 cycles will give a 103-fold ampliication, 20 cycles will
give a 106-fold ampliication, and 30 cycles will give a 109-fold ampliication.
THOUGHT PROBLEMS
8–59
Figure 8–11 shows a picture of DNA fragments that have been separated
by gel electrophoresis and then stained by ethidium bromide, a molecule that luoresces intensely under long-wavelength UV light when it is
bound to DNA. Such gels are a standard way of detecting the products
of cleavage by restriction nucleases. For the DNA fragment shown in
Figure 8–12, decide whether it will be cut by the restriction nucleases
EcoRI (5ʹ-GAATTC), AluI (5ʹ-AGCT), and PstI (5ʹ-CTGCAG). For those
that cut the DNA, how many products will be produced?
5′-AAGAATTGCGGAATTCGAGCTTAAGGGCCGCGCCGAAGCTTTAAA-3′
3′-TTCTTAACGCCTTAAGCTCGAATTCCCGGCGCGGCTTCGAAATTT-5′
Figure 8–12 A segment of double-stranded DNA (Problem 8–59).
8–60
he restriction nucleases BamHI and PstI cut their recognition sequences
as shown in Figure 8–13.
A. Indicate the 5ʹ and 3ʹ ends of the cut DNA molecules.
B. How would the ends be modiied if you incubated the cut molecules with
DNA polymerase in the presence of all four dNTPs?
C. After the reaction in part B, could you still join the BamHI ends together
by incubation with T4 DNA ligase? Could you still join the PstI ends
together? (T4 DNA ligase will join blunt ends together as well as cohesive
ends.)
D. Will joining of the ends in part C regenerate the BamHI site? Will it regenerate the PstI site?
8–61
he restriction nuclease EcoRI recognizes the sequence 5ʹ-GAATTC
and cleaves between the G and A to leave 5ʹ protruding single strands
(like BamHI, see Figure 8–13A). PstI, on the other hand, recognizes the
sequence 5ʹ-CTGCAG and cleaves between the A and G to leave 3ʹ protruding single strands (see Figure 8–13B). hese two recognition sites are
displayed on the helical representations of DNA in Figure 8–14.
A. For each restriction site, indicate the position of cleavage on each strand
of the DNA.
(B) PstI CLEAVAGE
(A) BamHI CLEAVAGE
3′
5′
-- G G A T C C --- C C T A G G -5′
3′
3′
5′
-- C T G C A G --- G A C G T C -5′
3′
BamHI
-- G
-- C C T A G
Figure 8–11 DNA fragments separated
by gel electrophoresis and stained
with ethidium bromide (Problem 8–59).
Each of the orange bands on the gel
represents a site where fragments of DNA
have migrated during electrophoresis.
Ethidium intercalates between base pairs
in double-stranded DNA. Removal of
ethidium from the aqueous environment
and ixing its orientation in the nonpolar
environment of DNA enhance its
luorescence dramatically. When
irradiated with long-wavelength UV light,
it luoresces a bright orange.
G A T C C -G --
PstI
-- C T G C A
-- G
G -A C G T C --
Figure 8–13 Restriction nuclease
cleavage of DNA (Problem 8–60).
(A) BamHI cleavage. (B) PstI cleavage.
only the nucleotides that form the
recognition sites are shown.
ANALYZING AND MANIPULATING DNA
179
B. From the positions of the cleavage sites, decide for each restriction
nuclease whether you expect it to approach the recognition site from the
major-groove side or from the minor-groove side.
8–62
3′
EcoRI
5′
Which, if any, of the restriction nucleases listed in Table 8–5 will
deinitely cleave a segment of cDNA that encodes the peptide KIGPACF?
(See Tables 7 and 8, page 966, for the genetic code.)
G
A
8–63
8–64
8–65
You wish to make a restriction map of a 3.0-kb BamHI restriction fragment. You digest three samples of the fragment with EcoRI, HpaII, and
a mixture of EcoRI and HpaII. You then separate the fragments by gel
electrophoresis and visualize the DNA bands by staining with ethidium
bromide (Figure 8–15). From these results, draw a restriction map that
shows the relative positions of the EcoRI and HpaII recognition sites and
the distances in kilobases (kb) between them.
A
minor
groove
T
T
C
5′
If you add DNA to wells at the top of a gel, should you place the positive electrode (anode) at the top or at the bottom of the gel? Explain your
choice.
You want to clone a DNA fragment that has KpnI ends into a vector that
has BamHI ends. he problem is that BamHI and KpnI ends are not
compatible: BamHI leaves a 5ʹ overhang and KpnI leaves a 3ʹ overhang
(Figure 8–16). A friend suggests that you try to link them with an oligonucleotide “splint” as shown in Figure 8–16. It is not immediately clear
to you that such a scheme will work because ligation requires an adjacent 5ʹ phosphate and 3ʹ hydroxyl. Although molecules that are cleaved
with restriction nucleases have appropriate ends, oligonucleotides are
typically synthesized with hydroxyl groups at both ends. Also, although
the junction shown in Figure 8–16 is BamHI–KpnI, the other junction is
KpnI–BamHI, and you are skeptical that the same oligonucleotide could
splint both junctions.
A. Draw a picture of the KpnI–BamHI junction and the oligonucleotide
splint that would be needed. Is this oligonucleotide the same or diferent
from the one shown in Figure 8–16?
B. Draw a picture of the molecule after treatment with DNA ligase. Indicate
which if any of the nicks will be ligated.
C. Will your friend’s scheme work?
8–66
How would a DNA sequencing reaction be afected if the ratio of dideoxynucleoside triphosphates (ddNTPs) to deoxynucleoside triphosphates
(dNTPs) were increased? What would the consequences be if the ratio
were decreased?
8–67
DNA sequencing of your own two β-globin genes (one from each of your
two copies of chromosome 11) reveals a mutation in one of the genes.
3′
PstI
5′
C
T
G
C
major
groove
A
G
5′
3′
Figure 8–14 Restriction sites on helical
DNA (Problem 8–61).
EcoRI
+
HpaII
HpaII
EcoRI
3′
TABLE 8–5 A set of restriction
nucleases and their recognition
sequences (Problem 8–62).
1.7 kb
Restriction
nuclease
Recognition
sequence
AluI
AGCT
Sau96I
GGNCC
HindIII
AAGCTT
N stands for any nucleotide.
1.6 kb
1.4 kb
0.9 kb
1.2 kb
0.9 kb
0.4 kb
0.5 kb
0.4 kb
Figure 8–15 Sizes of DNA bands
produced by digestion of a 3.0-kb
fragment by EcoRI, HpaII, and a mixture
of the two (Problem 8–63). Sizes of the
fragments are shown in kilobases.
Chapter 8: Analyzing Cells, Molecules, and Systems
180
BamHI-cut vector DNA
3′
DNA to be amplified
G
CCTAG
5′
GATCC
G
5′
3′
5′
C
CATGG
3′
GGTAC
C
5′
5′ -GACCTGTGGAAGC
3′ -CTGGACACCTTCG
primers
3′
KpnI-cut fragment
oligonucleotide “splint”
5′
3′
GATCGTAC
3′
5′
G
C
CCTAG CATGG
5′ 3′
BamHI
CATACGGGATTGA-3′
GTATGCCCTAACT-5′
Figure 8–16 Scheme to use
an oligonucleotide “splint” to
link incompatible restriction
ends (Problem 8–65).
(1) 5′ -GACCTGTGGAAGC-3′
(5) 5′ -CATACGGGATTGA-3′
(2) 5′ -CTGGACACCTTCG-3′
(6) 5′ -GTATGCCCTAACT-3′
(3) 5′ -CGAAGGTGTCCAG-3′
(7) 5′ -TGTTAGGGCATAC-3′
(4) 5′ -GCTTCCACAGGTC-3′
(8) 5′ -TCAATCCCGTATG-3′
Figure 8–17 DNA to be ampliied and potential PCR primers
(Problem 8–68).
KpnI
Given this information alone, how much should you worry about being a
carrier of an inherited disease that could be passed on to your children?
What other information would you like to have to assess your risk?
8–68
You want to amplify the DNA between the two stretches of sequence
shown in Figure 8–17. Of the listed primers, choose the pair that will
allow you to amplify the DNA by PCR.
8–69
In the very irst round of PCR using genomic DNA, the DNA primers
prime synthesis that terminates only when the cycle ends (or when a
random end of DNA is encountered). Yet, by the end of 20 to 30 cycles—
a typical ampliication—the only visible product is deined precisely by
the ends of the DNA primers (Figure 8–18). In what cycle is a doublestranded fragment of the correct size irst generated?
8–70
You want to express a rare human protein in bacteria so you can make
large quantities of it. To aid in its puriication, you decide to add a stretch
of six histidines to the N-terminus or the C-terminus of the protein. Such
histidine-tagged proteins bind tightly to Ni2+ columns but can be readily
eluted with a solution of EDTA or imidazole. his procedure allows an
enormous puriication in one step.
he nucleotide sequence that encodes your protein is shown in
Figure 8–19. Design a pair of PCR primers, each with 18 nucleotides of
homology to the gene, that will amplify the coding sequence and add
an initiation codon followed by six histidine codons to the N-terminus.
Design a pair of primers that will add six histidine codons followed by a
stop codon to the C-terminus.
FIRST CYCLE OF PCR
5′
3′
3′
5′
20 CYCLES
Figure 8–18 Products of PCR after 1 and
20 cycles (Problem 8–69).
ANALYZING AND MANIPULATING DNA
181
N-terminus
5′
H
H
GGT CGT ATG GCT ACT CGT CGC GCT GCT
M
A
T
R
R
A
A
C
H
CTT GCT GCA AGT CTC TCT TAG AAG TGT
L
A
A
S
L
S
*
C
N
3′
H
H
C
N
H
H
C
N
N
C
C
H
H
+ H
C-terminus
Figure 8–19 Nucleotide sequence around the N- and C-termini of the
protein you want to modify (Problem 8–70). The encoded amino acid
sequence is indicated below each codon using the one-letter code.
The asterisk (*) indicates the stop codon. only the top strand of the
double-stranded DNA is shown.
Figure 8–20 Structure of imidazole (Problem 8–71).
G A T
8–71
8–72
You have now cloned in an expression vector both versions of the histidine-tagged protein you created in Problem 8–70. Neither construct
expresses particularly strongly in bacteria, but the product is soluble. You
pass the crude extract over a Ni2+ ainity column, which binds histidinetagged proteins speciically. After washing the column extensively, you
elute your protein from the column using a solution containing imidazole (Figure 8–20), which releases your protein.
When you subject the eluted protein to electrophoresis and stain the
gel for protein, you are pleased to ind bands in the eluate that are not
present when control bacteria are treated similarly. But you are puzzled to see that the construct tagged at the N-terminus gives a ladder of
shorter proteins below the full-length protein, whereas the C-terminally
tagged construct yields exclusively the full-length protein. he amount of
full-length protein is about the same for each construct.
A. Why does a solution of imidazole release a histidine-tagged protein from
the Ni2+ column?
B. Ofer an explanation for the diference in the products generated by the
two constructs.
An example of a dideoxy sequencing gel is shown in Figure 8–21. Try
reading it. As read from the bottom of the gel to the top, the sequence corresponds to the mRNA for a protein. Can you ind the open reading frame
in this sequence? What protein does it code for?
C
100
50
CALCULATIONS
8–73
he restriction nuclease Sau3A recognizes the sequence 5ʹ-GATC and
cleaves on the 5ʹ side (to the left) of the G. (Since the top and bottom
strands of most restriction sites read the same in the 5ʹ-to-3ʹ direction,
only one strand of the site need be shown.) he single-stranded ends
produced by Sau3A cleavage are identical to those produced by BamHI
cleavage (see Figure 8–13), allowing the two types of ends to be joined
together by incubation with DNA ligase. (You may ind it helpful to draw
out the product of this ligation to convince yourself that it is true.)
A. What fraction of BamHI sites (5ʹ-GGATCC) can be cut with Sau3A? What
fraction of Sau3A sites can be cut with BamHI?
B. If two BamHI ends are ligated together, the resulting site can be cleaved
again by BamHI. he same is true for two Sau3A ends. Suppose you ligate
a Sau3A end to a BamHI end. Can the hybrid site be cut with Sau3A? Can
it be cut with BamHI?
C. What do you suppose is the average size of DNA fragments produced by
digestion of chromosomal DNA with Sau3A? What’s the average size with
BamHI?
Figure 8–21 A dideoxy sequencing gel of a cloned segment of DNA
(Problem 8–72). The lanes are labeled G, A, T, and C to indicate which
ddNTP was included in the reaction.
1
Chapter 8: Analyzing Cells, Molecules, and Systems
182
protein sequence
M
Q
K
F
N
A A T T
ATGCA AA TT AA
G G C C
degenerate oligonucleotide
Figure 8–22 A degenerate oligonucleotide probe for the Factor VIII
gene based on a stretch of amino acids from the protein (Problem
8–75). Because more than one DNA triplet can encode each amino
acid, a number of different nucleotide sequences are possible
for each amino acid sequence. During synthesis, a mixture of
nucleotides is included at each ambiguous position (for example,
A plus G at position 6). As a result, the mixture of synthesized
oligonucleotides contains all possible sequences that might encode
the speciic amino acid segment. Although only one of these
sequences in the genomic DNA will actually code for the protein, it
is impossible to tell in advance which one it is. The mixture of the
possible sequences—called a degenerate oligonucleotide probe—is
used to search a genomic library for the gene.
8–74
To prepare a genomic library, it is necessary to fragment the genome so
that it can be cloned in a vector. A common method is to use a restriction
nuclease.
A. How many diferent DNA fragments would you expect to obtain if you
cleaved human genomic DNA with Sau3A (5ʹ-GATC)? (Recall that there
are 3.2 × 109 base pairs in the haploid human genome.) How many would
you expect to get with EcoRI (5ʹ-GAATTC)?
B. Human genomic libraries are often made from fragments obtained by
cleaving human DNA with Sau3A in such a way that the DNA is only partially digested; that is, so that not all the Sau3A sites have been cleaved.
What is a possible reason for doing this?
8–75
A degenerate set of oligonucleotide probes for the Factor VIII gene for
blood clotting is shown in Figure 8–22. Each of these probes is only 15
nucleotides long. On average, how many exact matches to any single
15-nucleotide sequence would you expect to ind in the human genome
(3.2 × 109 base pairs)? How many matches to the collection of sequences
in the degenerate oligonucleotide probe would you expect to ind? How
might you determine that a match corresponds to the Factor VIII gene?
EcoRI
BamHI
4 kb
insert
vector
Figure 8–23 Recombinant plasmid
containing a cloned DNA segment
(Problem 8–76).
8–77
he DNA of certain animal viruses can integrate into a cell’s DNA
as shown schematically in Figure 8–25. You want to know the
ar
You have cloned a 4-kb segment of a gene into a plasmid vector (Figure
8–23) and now wish to prepare a restriction map of the gene in preparation for other DNA manipulations. Your advisor left instructions on how
to do it, but she is now on vacation, so you are on your own. You follow
her instructions, as outlined below.
1. Cut the plasmid with EcoRI.
2. Add a radioactive label to the EcoRI ends.
3. Cut the labeled DNA with BamHI.
4. Purify the insert away from the vector.
5. Digest the labeled insert briely with a restriction nuclease so that on
average each labeled molecule is cut about one time.
6. Repeat step 5 for several diferent restriction nucleases.
7. Run the partially digested samples side by side on an agarose gel.
8. Place the gel against x-ray ilm so that fragments with a radioactive
end can expose the ilm to produce an autoradiograph.
9. Draw the restriction map.
Your biggest problem thus far has been step 5; however, by decreasing
the amounts of nuclease and lowering the temperature, you were able to
ind conditions for partial digestion. You have now completed step 8, and
your autoradiograph is shown in Figure 8–24.
Unfortunately, your advisor was not explicit about how to construct a
map from the data in the autoradiograph. She is due back tomorrow. Will
you igure it out in time?
m
8–76
ke
rs
DATA HANDLING
kb
4.8
4.3
3.7
2.3
1.9
1.4
1.3
0.7
0.2
Figure 8–24 Autoradiograph showing the
electrophoretic separation of the labeled
fragments after partial digestion with the
three restriction nucleases represented by
the symbols (Problem 8–76). Numbers at
the left indicate the sizes of a set of marker
fragments in kilobases.
ANALYZING AND MANIPULATING DNA
183
Figure 8–25 Integration of viral DNA into
cell DNA (Problem 8–77).
viral DNA
(A) MAP OF VIRAL DNA
EcoRI
HpaII
cell DNA
e
a
d
BglI
b
c
integrated
viral DNA
BglI
structure of the viral genome as it exists in the integrated state. You digest
samples of viral DNA and DNA from cells that contain the integrated
virus with restriction nucleases that cut the viral DNA at known sites
(Figure 8–26A). Subsequently, you separate the fragments by electrophoresis on agarose gels and visualize the bands that contain viral DNA
by hybridization with a viral DNA probe. You obtain the patterns shown
in Figure 8–26B.
From this information, decide in which of the ive segments of the viral
genome (labeled a to e in Figure 8–26A) the integration event occurred.
HpaII
(B) RESTRICTION DIGESTS
EcoRI
virus cell
HpaII
virus cell
BglI
virus cell
MEDICAL LINKS
8–78
Many mutations that cause human genetic diseases involve the substitution of one nucleotide for another, as is the case for sickle-cell anemia
(Figure 8–27A). An assay based on ligation of oligonucleotides provides
a rapid way to detect such speciic single-nucleotide diferences. his
assay uses pairs of oligonucleotides: for each pair, one oligonucleotide
is labeled with biotin and the other with a radioactive (or luorescent)
tag. In the assay shown in Figure 8–27B for the detection of the mutation responsible for sickle-cell anemia, two pairs of oligonucleotides are
hybridized to DNA from an individual and incubated in the presence of
DNA ligase. Biotinylated oligonucleotides are then bound to streptavidin
on a solid support and any associated radioactivity is visualized by autoradiography, as shown in Figure 8–27C.
A. Do you expect the βA and βS oligonucleotides to hybridize to both βA and
βS DNA?
B. How does this assay distinguish between βA and βS DNA?
Figure 8–26 Viral and cell DNA digested
with various restriction nucleases and
hybridized to a viral DNA probe (Problem
8–77). (A) Restriction sites on the viral
genome. The DNA segments deined by
these sites are indicated by the letters a
to e. (B) Restriction digests of viral DNA
and cellular DNA. Agarose gels separate
DNA fragments on the basis of size—the
smaller the fragment, the farther it moves
toward the bottom of the gel.
(A) β-GLOBIN SEQUENCE
normal (β A) sequence
A
ATGGTGCACCTGACTCCTG
GGAGAAGGTCTGCCGTTACTG
T
sickle-cell mutant (β S) sequence
(B)
OLIGONUCLEOTIDES
1
β A oligo
biotin-ATGGTGCACCTGACTCCTGA
β S oligo
2
(C)
radioactive oligo
32P-GGAGAAGGTCTGCCGTTACTG
biotin-ATGGTGCACCTGACTCCTGT
ASSAY
β Aβ A homozygote
β Aβ S heterozygote
β Sβ S homozygote
oligos
oligos
1 + 3
2 + 3
3
Figure 8–27 oligonucleotide-ligation
assay (Problem 8–78). (A) Sequence of
the β-globin gene around the site of the
sickle-cell (βS) mutation. The normal βA
sequence carries an A at the central
position; the sickle-cell mutant βS
sequence has a T instead. (B) Speciic
oligonucleotides for ligation assay.
(C) Assays to detect the single-nucleotide
difference between the βA and βS
sequences. After hybridization to patient
DNA, biotinylated oligonucleotides were
collected in a spot on a sheet of ilter
paper and exposed to x-ray ilm to detect
radioactivity, which turns the ilm black.
184
Chapter 8: Analyzing Cells, Molecules, and Systems
(A) DMD GENE
a
0
b
c
d e
0.5
f
g
1.0
h
i
sites amplified
by PCR
1.5
2.0 (Mb)
(B) MULTIPLEX PCR ANALYSIS
DMD patients
g
h
e
d
i
b
c
f
normal A
B
C
D
E
F
0
a
8–79
Duchenne’s muscular dystrophy (DMD) is among the most common
human genetic diseases, afecting approximately 1 in 3500 male births.
One-third of all new cases arise via new mutations. he DMD gene, which
is located on the X chromosome, is greater than 2 million base pairs in
length and contains at least 70 exons. Large deletions account for about
60% of all cases of the disease, and they tend to be concentrated around
two regions of the gene.
he very large size of the DMD gene complicates the analysis of mutations. One rapid approach, which can detect about 80% of all deletions,
is termed multiplex PCR. It uses multiple pairs of PCR primers to amplify
nine diferent segments of the gene in the two most common regions for
deletions (Figure 8–28A). By arranging the PCR primers so that each pair
gives a diferent size product, it is possible to amplify and analyze all nine
segments in one PCR reaction. An example of multiplex PCR analysis of
six unrelated DMD males is shown in Figure 8–28B.
A. Describe the extent of the deletions, if any, in each of the six DMD
patients.
B. What additional control might you suggest to conirm your analysis of
patient F?
STUDYING GENE EXPRESSION AND FUNCTION
TERMS TO LEARN
allele
chromatin immunoprecipitation
complementation test
conditional mutation
DNA microarray
epistasis analysis
genetic screen
genetics
genotype
green fluorescent protein (GFP)
haplotype block
phenotype
polymorphism
quantitative RT-PCR
reverse genetics
single-nucleotide polymorphism (SNP)
transgene
transgenic organism
DEFINITIONS
Match each deinition below with its term from the list above.
8–80
A search through a large collection of mutants for a mutant with a particular phenotype.
8–81
One of a number of common sequence variants that coexist in the population.
Figure 8–28 Multiplex PCR analysis
of six DMD patients (Problem 8–79).
(A) The DMD gene with the nine sites
ampliied by PCR indicated by arrows.
The sizes of the PCR products are so
small on this scale that their location is
simply indicated. (B) Agarose gel display
of ampliied PCR products. “Normal”
indicates a normal male. The lane marked
“0” shows a negative control with no
added DNA.
STUDYING GENE EXPRESSION AND FUNCTION
8–82
One of a set of alternative forms of a gene.
8–83
he observable character of a cell or an organism.
8–84
Comparing the phenotypes of diferent combinations of mutations to
determine the order in which the genes act.
8–85
Ancestral chromosome segment that has been inherited with little
genetic rearrangement across generations.
8–86
Animal or plant that has been permanently engineered by gene deletion,
gene insertion, or gene replacement.
8–87
he genetic constitution of an individual cell or organism.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
8–88
In an organism whose genome has been sequenced, identifying the
mutant gene responsible for an interesting phenotype is as easy for
mutations induced by chemical mutagenesis as it is for those generated
by insertional mutagenesis.
8–89
Loss-of-function mutations are usually recessive.
8–90
If two mutations have a synthetic phenotype, it usually means that the
mutations are in genes whose products operate in the same pathway.
THOUGHT PROBLEMS
8–91
Distinguish between the following genetic terms:
A. Locus and allele
B. Homozygous and heterozygous
C. Genotype and phenotype
D. Dominant and recessive
8–92
Explain the diference between a gain-of-function mutation and a dominant-negative mutation. Why are both these types of mutation usually
dominant?
8–93
Discuss the following statement: “We would have no idea today of the
importance of insulin as a regulatory hormone if its absence were not
associated with the human disease diabetes. It is the dramatic consequences of its absence that focused early eforts on the identiication of
insulin and the study of its normal role in physiology.”
8–94
What are single-nucleotide polymorphisms (SNPs), and how can they be
used to locate a mutant gene?
8–95
Fatty acid synthase in mammalian cells is encoded by a single gene. his
remarkable protein carries out seven distinct biochemical reactions. he
mammalian fatty acid synthase gene is homologous to seven diferent
E. coli genes, each of which encodes one of the functions of the mammalian protein. Do you think it is likely that the proteins in E. coli function
together as a complex? Why or why not?
8–96
How does reverse genetics difer from standard genetics?
8–97
he cells in an individual animal contain nearly identical genomes. In
an experiment, a tissue composed of multiple cell types is ixed and subjected to in situ hybridization with a DNA probe to a particular gene. To
your surprise, the hybridization signal is much stronger in some cells
than in others. Explain this result.
185
186
Chapter 8: Analyzing Cells, Molecules, and Systems
Figure 8–29 Site-directed mutagenesis
(Problem 8–98). (A) Sequence of DNA and
the encoded protein. (B) Conversion of
normal gene into a mutant gene.
(A)
L R D P Q G G V I
5′ -CTTAGAGACCCGCAGGGCGGCGTCATC- 3′
(B)
your gene
CAG
GCC
mutator
oligo
DNA POLYMERASE,
DNA LIGASE
CAG
G C
C
INTRODUCE DNA INTO CELLS
AND CLONE RESULTING VIRUSES
CAG
GTC
C GG
GCC
wild-type gene
mutant gene (Q → R)
8–98
From previous work, you suspect that the glutamine (Q) in the protein
segment in Figure 8–29A plays an important role at the active site. Your
advisor wants you to alter the protein in three ways: change the glutamine to arginine (R) change the glutamine to glycine (G), and delete
the glutamine from the protein. You plan to accomplish these mutational
alterations on a version of your gene that is cloned into M13 viral DNA.
You want to hybridize an appropriate oligonucleotide to the M13 viral
DNA, so that when DNA polymerase extends the oligonucleotide around
the single-stranded M13 circle, it will complete a strand that encodes the
complement of the desired mutant protein. Design three 20-nucleotidelong oligonucleotides that could be hybridized to the cloned gene on single-stranded M13 viral DNA as the irst step in efecting the mutational
changes (Figure 8–29B).
8–99
You have just gotten back the results from an RNA-seq analysis of mRNA
from liver. You had anticipated counting the number of reads of each
mRNA to determine the relative abundance of diferent mRNAs. But you
are puzzled because many of the mRNAs have given you results like those
shown in Figure 8–30. How is it that diferent parts of an mRNA can be
represented at diferent levels?
reads
mRNA
exons
1
2
3
4
5
Figure 8–30 RNA-seq reads for a liver
mRNA (Problem 8–99). The exon structure
of the mRNA is indicated, with proteincoding segments indicated in light blue
and untranslated regions in dark blue.
The numbers of sequencing reads are
indicated by the heights of the vertical
lines above the mRNA.
STUDYING GENE EXPRESSION AND FUNCTION
187
brick-red
white
DATA HANDLING
Early genetic studies in Drosophila laid the foundation for our current
understanding of genes. Drosophila geneticists were able to generate
mutant lies with a variety of easily observable phenotypic changes.
Alterations from the ly’s normal brick-red eye color have a venerable
history because the very irst mutant found by homas Hunt Morgan
was a white-eyed ly (Figure 8–31). Since that time, a large number of
mutant lies with intermediate eye colors have been isolated and given
names that challenge your color sense: garnet, ruby, vermilion, cherry,
coral, apricot, buf, and carnation. he mutations responsible for these
eye-color phenotypes are recessive. To determine whether the mutations
afected the same or diferent genes, lies homozygous for each mutation were bred to one another in pairs and the eye colors of their progeny
were noted. In Table 8–6, brick-red wild-type eyes are shown as (+) and
other colors are indicated as (–).
A. How is it that lies with two diferent eye colors—ruby and white, for
example—give rise to progeny that all have brick-red eyes?
B. Which mutations afect diferent genes and which mutations are alleles
of the same gene?
C. How can alleles of the same gene give diferent eye colors? hat is to say,
why don’t all the mutations in the same gene give the same phenotype?
8–100
8–101
Figure 8–31 Drosophila with different
color eyes (Problem 8–100). Wild-type
lies with brick-red eyes are shown on the
left and white-eyed lies are shown on the
right. Flies with eye colors between red
and white are shown in between.
You have designed and constructed a DNA microarray that carries 20,000
allele-speciic oligonucleotides (ASOs). hese ASOs correspond to the
wild-type and mutant alleles associated with 1000 human diseases. You
have designed the microarray so the ASO that hybridizes to the mutant
allele is located right below the ASO that hybridizes to the same site in
the wild-type sequence. his arrangement is illustrated in Figure 8–32 for
ASOs that are speciic for the sickle-cell allele (βS) and the corresponding
TABLE 8–6 Complementation analysis of Drosophila eye-color mutations (Problem 8–100).
MUTATIoN
White
Garnet
Ruby
Vermilion
white
garnet
ruby
vermilion
cherry
coral
apricot
buff
carnation
–
+
+
+
–
–
–
–
+
–
+
+
+
+
+
+
+
–
+
+
+
+
+
+
–
+
+
+
+
+
–
–
–
–
+
–
–
–
+
–
–
+
–
+
Cherry
Coral
Apricot
Buff
Carnation
Brick-red eyes are indicated as (+). Other colors are indicated as (–).
–
188
Chapter 8: Analyzing Cells, Molecules, and Systems
(A) β-GLOBIN ALLELES
(B) DNA MICROARRAY
ASOβA
βA
ASOβS
βS
sickle-cell
mutation
ASOβA
ASOβS
site in the wild-type allele (βA). ASOβS hybridizes to the sickle-cell mutation, and ASOβA hybridizes to the corresponding position in the wild-type
allele. As a test of your microarray, you carry out hybridizations of DNA
isolated from individuals who are homozygous for the wild-type allele,
homozygous for the sickle-cell allele, or heterozygous for the wild-type
and sickle-cell alleles. For each DNA sample, draw the expected patterns
of hybridization to the globin ASOs on your microarray.
8–102
Now that news of your disease-speciic DNA microarray has gotten
around, you are being inundated with requests to analyze various samples. Just today you received requests from four physicians for help in
the prenatal diagnosis of the same disease. Each of the pregnant mothers
has a family history of this disease. You included on your array the ive
alleles known to cause this disease (Figure 8–33). Each of these alleles
is recessive. You agree to help. You prepare samples of fetal DNA gotten
by amniocentesis and hybridize them to your microarrays. Your data
are shown in Figure 8–33C. Assuming that the ive disease alleles shown
in Figure 8–33A are the only ones in the human population, decide for
each sample of DNA whether the individual will have the disease or not.
Explain your reasoning.
8–103
You’ve heard about this cool technique for targeted recombination into
embryonic stem (ES) cells that allows you to create a null allele or a conditional allele more or less at the same time. As your friend explained it
to you, you irst carry out standard gene targeting into ES cells by homologous recombination, as shown in Figure 8–34. he Neo gene, which
codes for resistance to the antibiotic neomycin, allows selection for ES
cells that have incorporated the vector. hese cells can then be screened
by Southern blotting for those that have undergone a targeted event.
he really cool part is to lank the Neo gene and an adjacent exon or two
(A) DISEASE ALLELES
1
2
3
4
(B) DNA MICROARRAY
5
wild type
m1
mutant 1
m2
mutant 2
m3
mutant 3
1
m4
mutant 4
5
wild-type ASOs
m1 m2 m3 m4 m5
2
mutant ASOs
m5
mutant 5
(C) HYBRIDIZATION DATA
JC
Figure 8–32 DNA microarray for detection
of disease alleles (Problem 8–101).
(A) The β-globin alleles. The wild-type
β-globin gene (βA) and the sickle-cell
allele (βS) are shown. The position of the
sickle-cell mutation is shown by a vertical
line. The ASos are shown as short red
lines arranged above the sites in the
gene to which they hybridize. Because
the ASos are located at corresponding
positions, each is speciic for its allele:
the wild-type ASo will not hybridize to
the sickle-cell allele, nor will the sicklecell ASo hybridize to the wild-type allele.
(B) DNA microarray. A tiny section of
the microarray is enlarged to illustrate the
locations of the wild-type and sickle-cell
ASos.
BF
HK
TW
3
4
Figure 8–33 DNA microarray analysis
of alleles present in prenatal samples
(Problem 8–102). (A) Wild-type and
disease alleles. Vertical lines indicate the
sites of the mutations in the diseasecausing alleles. The ASos speciic for
the disease mutations are shown as
red lines and labeled m1, m2, etc. The
corresponding ASos that hybridize to the
wild-type gene at sites that correspond
to the position of the mutations are
labeled 1, 2, etc. (B) DNA microarray. The
arrangement of wild-type and mutant
ASos is indicated. (C) Hybridization data.
Samples of fetal DNA were hybridized
to DNA arrays. Dark spots indicate sites
where hybridization occurred. Letters
identify the patients.
STUDYING GENE EXPRESSION AND FUNCTION
lox
189
lox
2
lox
Neo
3
4
+
1
2
3
4
5
4
5
HOMOLOGOUS
RECOMBINATION
lox
1
lox
2
3
lox
Neo
Figure 8–34 Targeted modiication of a
gene in mouse ES cells (Problem 8–103).
Exons are shown as boxes; the promoter
is identiied by the horizontal arrow. The
targeting vector is fully homologous to
the gene, except for the presence of the
three lox sites and the Neo gene, all of
which are in introns. Cre recombinase,
which can be introduced by transfection
of a Cre-expression vector, catalyzes
a site-speciic recombination event
between a pair of lox sites in about
20% of transfected cells.
Cre RECOMBINASE
with lox sites. his technique is commonly referred to as “loxing.” Once
the modiied ES cells have been identiied, they can be exposed to the
Cre recombinase, which promotes site-speciic recombination between
pairs of lox sites. One advantage is that it allows you to get rid of the Neo
gene and any bacterial DNA segments, which can sometimes inluence
the phenotype.
A. What possible products might you get from expression of Cre in modiied
ES cells that carry three lox sites, as indicated in Figure 8–34?
B. Which product(s) would be a null allele?
C. Which product(s) would have a pair of lox sites but be an otherwise normal allele?
D. If you had one mouse that expressed Cre under the control of a tissuespeciic promoter, can you use the allele in part C (after you’ve put it into
the germ line of a mouse) as a conditional allele; that is, one whose defect
is expressed only in a particular tissue?
CRISPR/Cas9–guide RNA complexes hold enormous promise as aids for
genome engineering in plants and animals. he CRISPR system is almost
too good to be true. You want to test just how speciic the Cas9–guide
RNA complexes are; that is, whether they really recognize individual sites
in the genome, which is the basis for their touted actions. You realize
that you can test their speciicity using the “DNA curtain” assay you have
developed. You make the DNA curtain by tethering single molecules of
bacteriophage lambda DNA (about 50,000 nucleotides) at one end, and
then stretching them in the same direction by lowing bufer across the
slide. You incubate the DNA curtain with a highly luorescent version of
Cas9—either loaded with the guide RNA or free—and visualize the distribution of the Cas9 by sensitive luorescence microscopy, as shown in
Figure 8–35.
A. Phage lambda DNA has a single site that perfectly matches the guide
RNA. Does your experiment support the existence of a single site that is
recognized by the Cas9–guide RNA complex? Explain your answer.
8–104
(A)
RNA-loaded Cas9
λ2
5 μm
(B)
RNA-free Cas9
5 μm
Figure 8–35 DNA curtain assay for target
binding by Cas9–guide RNA (Problem
8–104). (A) DNA curtain incubated with
Cas9–guide RNA that matches one site in
the lambda genome. Pink spots indicate
the sites where Cas9–guide RNA is
located. (B) DNA curtains incubated with
Cas9 in the absence of guide RNA.
Chapter 8: Analyzing Cells, Molecules, and Systems
190
B. One of the main concerns for using the CRISPR system is that the Cas9–
guide RNA will bind to other sites that are related to the intended target
(so-called of-target efects). Is there any evidence for of-target binding
in your experiment? Do you think the results would be any diferent if you
had a DNA curtain made from human chromosome 1 (about 250,000,000
nucleotides)?
MATHEMATICAL ANALYSIS OF CELL FUNCTIONS
TERMS TO LEARN
robustness
stochastic
DEFINITIONS
Match each deinition below with its term from the list above.
8–105
he ability of biological regulatory systems to function normally in the
face of frequent and sometimes extreme variations in external conditions
or the concentrations or activities of key components.
8–106
Describes random variations in protein content of individual cells resulting in variations in cell phenotypes.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
8–107
To judge the biological importance of an interaction between protein A
and protein B, we need to know quantitative details about their concentrations, ainities, and kinetic behaviors.
8–108
he association constant, Ka, is equal to 1 minus the dissociation constant, Kd; that is, Ka = 1 – Kd.
8–109
he rate of change in the concentration of any molecular species X is
given by the balance between its rate of appearance and its rate of disappearance.
8–110
After a sudden increase in transcription, a protein with a slow rate of degradation will reach a new steady-state level more quickly than a protein
with a rapid rate of degradation.
8–111
he Lac operon, which is turned on by a transcription activator and
turned of by a transcription repressor, is a classic example of an AND
NOT logic gate.
THOUGHT PROBLEMS
8–112
Consider the two network motifs illustrated in Figure 8–36. Both contain
only negative regulation by repressors, and yet one is a negative feedback
(B)
ACTIVATING
INPUT
(A)
ACTIVATING
INPUT
GENE X
GENE Y
GENE X
GENE Y
GENE Z
X
Y
X
Y
Z
Figure 8–36 Two network motifs
composed of transcriptional repressors
(Problem 8–112). (A) A two-gene network.
(B) A three-gene network.
MATHEMATICAL ANALYSIS OF CELL FUNCTIONS
(A)
ACTIVATING
INPUT
191
(B)
ACTIVATING
INPUT
(C)
ACTIVATING
INPUT
(D)
ACTIVATING
INPUT
GENE X
GENE X
GENE X
GENE X
X
X
X
X
GENE Y
GENE Y
GENE Y
GENE Y
Y
Y
Y
Y
GENE Z
GENE Z
GENE Z
GENE Z
Z
Z
Z
Z
loop and the other is a positive feedback loop. Which is the negative and
which is the positive feedback loop? Explain how networks made up of
negative regulatory elements can difer in their behavior.
8–113
Examine the network motifs in Figure 8–37. Decide which ones are negative feedback loops and which are positive. Explain your reasoning.
8–114
Imagine that a random perturbation positions a bistable system
precisely at the boundary between two stable states (at the orange dot in
Figure 8–38). How would the system respond?
8–115
For the network motif shown in Figure 8–39A, decide whether the pulse
of protein X output (Figure 8–39B) requires that the equilibrium constant
for the binding of transcription activator A (KA) must be much greater or
much less than the equilibrium constant for binding of the transcription
repressor R (KR). Explain your reasoning.
1
2
(B)
SUDDEN
ACTIVATING
INPUT
A
A
inactive
33
A
concentration of X
fast gene A
activation
R slow gene
repression
rate of protein synthesis
concentration of Y
(A)
Figure 8–37 Network motifs composed
of transcription activators and repressors
(Problem 8–113).
sudden
activating
input
protein X
GENE X
Figure 8–38 Perturbations of a bistable
system (Problem 8–114). As shown by the
green lines, after perturbation 1 the system
returns to its original stable state (green dot
at left), and after perturbation 2, the system
moves to the other stable state (green dot at
right). Perturbation 3 moves the system to
the precise boundary between the two stable
states (orange dot).
time
X
Figure 8–39 Network motif that generates a pulse of protein X (Problem
8–115). (A) Regulation of gene X by a transcription activator (green circle) and
a transcription repressor (red circle). (B) Pulse of protein X in response to an
activating input.
Chapter 8: Analyzing Cells, Molecules, and Systems
192
CALCULATIONS
8–116
Consider the situation in which a transcription regulator (R) binds to the
promoter (pX) for gene X. If the concentration of a transcription regulator is 100 times 1/K, where K is the equilibrium constant for the binding
of the regulator to the promoter, what percentage of the promoters in a
population of cells will be occupied? What percentage of promoters will
be bound if [R] is equal to 1/K? If [R] is 100-fold less than 1/K? (he equation for bound fraction is [R:pX]/[pXT ] = K[R]/(1 + K[R]), where [R:pX] is the
complex of the regulator with the promoter and [pXT ] is the total concentration of promoters.)
8–117
What is the molar concentration of a promoter sequence that is present
at one copy per E. coli? Assume E. coli is a rod that is 0.8 μm in diameter
and 3 μm in length. (he equation for the volume of a cylinder is πr2h.)
8–118
If the concentration of a transcription repressor is 100 molecules per
E. coli, and it has a single binding site in the bacterial genome, what must
the equilibrium constant be for the binding site to be 99% occupied?
he Lac repressor regulates the expression of a set of genes for lactose
metabolism, which are adjacent to its binding site on the bacterial chromosome. In the absence of lactose in the medium, the binding of the
repressor turns the genes of. When lactose is added, an inducer is generated that binds to the repressor and prevents it from binding to its DNA
target, thereby turning on gene expression.
Inside E. coli there are about 10 molecules of Lac repressor (10–8 M)
and 1 binding site (10–9 M) on the bacterial genome. he equilibrium
constant, K, for binding of the repressor to its binding site is 1013 M–1. In
the presence of lactose, when an inducer of gene expression binds to the
repressor, K for repressor binding to its DNA binding sites decreases to
1010 M–1.
A. In a population of bacteria growing in the absence of lactose, what fraction of the binding sites would you expect to be bound by repressor?
B. In bacteria growing in the presence of lactose, what fraction of binding
sites would you expect to be bound by the repressor?
C. Given the information in this problem, would you expect the inducer to
turn on gene expression? Why or why not?
D. he Lac repressor binds nonspeciically to any sequence of DNA with a K
of about 106 M–1, which is a very low ainity. Can you suggest in a qualitative way how such low-ainity, nonspeciic binding might alter the calculations in parts A and B and your conclusion in part C?
8–119
8–120
Equilibrium dialysis provides a simple method for determining the equilibrium constant for binding of a ligand (L) by a protein (Pr). he protein
is conined inside a dialysis sac, formed by an artiicial membrane with
pores too tiny for the protein to enter, but which the much smaller ligand
can freely permeate. he ligand, usually radiolabeled, is added to the
solution surrounding the dialysis sac; after equilibrium has been established, the concentration of the ligand is measured in both compartments. he concentration in the external compartment is the concentration of free (unbound) ligand and the concentration in the dialysis sac is
the sum of the bound (Pr–L) plus free ligand. By measuring these values
after various initial ligand concentrations, the value of the equilibrium
constant can be determined. By convention, the equilibrium is usually
considered for the dissociation reaction (Pr–L → Pr + L), rather than the
association reaction (Pr + L → Pr–L), and thus the equilibrium constant
in this case is referred to as the dissociation constant, Kd.
It is useful to look at the transformation of the standard equilibrium
relationship into the form commonly used to analyze the data. At equilibrium,
MATHEMATICAL ANALYSIS OF CELL FUNCTIONS
[Pr][L]
[Pr–L]
1.0
Given that the total protein concentration, [Pr]TOT, is the sum of the concentration of bound protein [Pr–L] and free protein [Pr], we can substitute [Pr]TOT – [Pr–L] for [Pr] and rearrange to give
[Pr–L] =
[Pr]TOT[L]
Kd + [L]
his is an equation for a rectangular hyperbola. It can be rearranged to
give a linear form, as was irst done by George Scatchard in 1947 (hence,
graphs of such data are commonly known as Scatchard plots):
bound/free
Kd =
193
0.5
A
0
B
2
4
6
IPTG bound (× 10–7 M)
Figure 8–40 Scatchard plot of equilibrium
dialysis data for the binding of IPTG to
the Lac repressor (Problem 8–120).
[Pr–L] – [Pr–L] [Pr]TOT
=
+
Kd
[L]
Kd
At constant protein concentration and a variety of ligand concentrations, a plot of bound over free ligand ([Pr–L]/[L]) against bound ligand
([Pr–L]) gives a line with slope equal to –1/Kd and an x-intercept equal to
[Pr]TOT, which is the total concentration of binding sites.
In the early 1960s, when the nature of the genetically identiied
repressors of bacterial gene expression had not been deined, Walter
Gilbert and Benno Müller-Hill used equilibrium dialysis to measure the
binding of an inducer of gene expression (IPTG) to the Lac repressor
protein. hey used radiolabeled IPTG and two partially puriied preparations of Lac repressor protein: one from wild-type cells and the other
from mutant cells in which induction of the lactose operon occurred at
lower concentrations of IPTG. he mutant cells were assumed to carry a
Lac repressor that bound IPTG more tightly. heir data are shown as a
Scatchard plot in Figure 8–40.
A. What are the Kd values for the two lines shown in the igure?
B. Which line corresponds to the wild-type Lac repressor and which to the
mutant (tighter IPTG-binding) repressor?
DATA HANDLING
8–121
Detailed analysis of the regulatory region of the Lac operon has revealed
surprising complexity. Instead of a single binding site for the Lac repressor, as might be expected, there are three sites termed operators: O1, O2,
and O3, arrayed along the DNA as shown in Figure 8–41. To probe the
functions of these three sites, you make a series of constructs in which
various combinations of operator sites are present. You examine their
92 bp
401 bp
1
O3
O1
2
O3
O1
3
O1
4
O1
5
O3
6
O3
7
8
2-mer 4-mer
O2
O2
O2
O2
110
6700
90
3900
80
1400
60
140
1
5
1
2
1
1
1
1
Figure 8–41 Repression of
β-galactosidase by promoter regions that
contain different combinations of Lac
repressor binding sites (Problem 8–121).
The base-pair (bp) separation of the
three operator sites is shown. Numbers
at right refer to the level of repression,
with higher numbers indicating more
effective repression by dimeric (2-mer) or
tetrameric (4-mer) repressors.
Chapter 8: Analyzing Cells, Molecules, and Systems
194
ability to repress expression of β-galactosidase, using either tetrameric
(wild type) or dimeric (mutant) forms of the Lac repressor. he dimeric
form of the repressor can bind to a single operator (with the same ainity
as the tetramer) with each monomer binding to half the site. he tetramer,
the form normally expressed in cells, can bind to two sites simultaneously. When you measure repression of β-galactosidase expression, you
ind the results shown in Figure 8–41, with higher numbers indicating
more efective repression.
A. Which single operator site is the most important for repression? How can
you tell?
B. Do combinations of operator sites (Figure 8–41, constructs 1, 2, 3, and 5)
substantially increase repression by the dimeric repressor? Do combinations of operator sites substantially increase repression by the tetrameric
repressor? If the two repressors behave diferently, ofer an explanation
for the diference.
C. he wild-type repressor binds O3 very weakly when it is by itself on a segment of DNA. However, if O1 is included on the same segment of DNA,
the repressor binds O3 quite well. How can that be?
MCAT STYLE
Passage 1 (Questions 8–122 to 8–126)
Cancer is caused by aberrant versions of our own genes. Chronic myelogenous
leukemia, for example, is caused by a hybrid chromosome—the Philadelphia
chromosome—that is formed by fusion of broken segments of chromosomes 9
and 22. By chance, this fusion links the Bcr gene from chromosome 22 to the Abl
gene from chromosome 9. he Bcr-Abl fusion gene produces a chimeric protein
that combines the protein kinase activity of Abl with an activating segment of Bcr
to produce a hyperactive Bcr-Abl fusion protein that drives cells of the immune
system to proliferate excessively.
Drugs that speciically inhibit such hyperactive kinases are revolutionizing
cancer treatment. he irst example of such a drug—imatinib—was discovered by
searching for compounds that bind and inhibit the Bcr-Abl protein kinase. Imatinib caused rapid disappearance of cancer cells with minimal side efects. Typically, however, after a few months or years, the cancer returns, this time in a form
that is resistant to the drug. he appearance of drug-resistant cancers is a major
challenge for many inhibitor-based therapies.
Which of the following techniques would serve best as the basis for a
rapid, highly sensitive blood test to detect circulating cancer cells that
carry the Bcr-Abl gene?
A. DNA sequencing
B. Flow cytometry
C. PCR analysis
D. Western blotting
8–122
You hypothesize that drug resistance is caused by mutations in the fusion
gene that block binding of imatinib to the Bcr-Abl protein. What would
be your irst step in testing this hypothesis?
A. Amplify the Bcr-Abl gene by PCR and sequence it to look for mutations in
the gene.
B. Chemically modify imatinib to search for new drugs that will kill the
resistant cells.
C. Determine whether the resistant cancer cells carry the Philadelphia
chromosome.
D. Purify Bcr-Abl protein from resistant cancer cells and test whether it
binds imatinib.
8–123
8–124
To test whether imatinib inhibits the activity of the Bcr-Abl kinases from
imatinib-resistant cancer cells from patients, you will need a lot of the
MCAT STYLE
Bcr-Abl protein. What would be the most rapid and direct way to obtain a
large amount of the protein?
A. Clone the Bcr-Abl gene from a genomic DNA library into an expression
vector and purify the protein after expression in bacteria.
B. Grow large amounts of drug-resistant cancer cells in culture and isolate
the protein after SDS polyacrylamide-gel electrophoresis.
C. Grow large amounts of drug-resistant cancer cells in culture and purify
the Bcr-Abl protein by column chromatography.
D. Use PCR to amplify the Bcr-Abl gene from cDNA, clone it into an expression vector, express the protein in bacteria, and then purify it.
As an alternative hypothesis, you consider that imatinib resistance arises
from mutations outside the coding region of the Bcr-Abl gene that cause
the gene to be expressed at abnormally high levels, thereby reducing the
efectiveness of imatinib. Which one of the following techniques would
you use to most rapidly test this hypothesis?
A. cDNA library analysis
B. In situ hybridization
C. Quantitative RT-PCR
D. Western blotting
8–125
If you ind that there are no mutations in the Bcr-Abl coding region and
the gene is not overexpressed, what would you do next to deine the basis
for imatinib resistance?
A. Look for mutations in regions neighboring the Bcr-Abl gene.
B. Sequence the genomic DNA from imatinib-resistant cells.
C. Use genetic analysis to identify the genes responsible for resistance.
D. Use genome-wide association studies to identify resistance genes.
8–126
Passage 2 (Questions 8–127 to 8–131)
Highly regulated proteolysis of cytoplasmic proteins plays an important role in
many cellular events. Biochemical analysis uncovered the molecular machinery
responsible for regulated proteolysis. Early studies established that damaged proteins added to cell lysates were rapidly destroyed in an ATP-dependent manner.
Subsequently, it was shown that a small protein called ubiquitin was covalently
attached to proteins before they were destroyed. Column chromatography was
used to purify the proteins that attach ubiquitin to other proteins. hese studies
led to the discovery of a biochemical pathway in which ubiquitin is irst activated
via formation of a high-energy thioester bond with a protein called E1. E1 then
passes the ubiquitin to another protein called E2. Finally, ubiquitin is covalently
attached to the target protein. A multiprotein complex called E3 binds to both E2
and the target protein, which ensures that the ubiquitin is attached to the correct
target.
Components of the ubiquitin machinery were independently identiied in
genetic screens for genes that regulate the cell cycle. hese screens were based on
the hypothesis that mutations in genes necessary for speciic steps in the cell cycle
should cause cells to arrest before completion of those steps. hus, a mutation in a
gene required for chromosome segregation should cause cells to arrest in mitosis
before chromosomes have separated. Mutagenized yeast cells were screened for
mutations that cause cells to arrest at speciic points in the cell cycle, which led
to a collection of cell-division control (Cdc) mutants. hree of these mutants—
Cdc16, Cdc23, and Cdc27—arrested in mitosis before chromosome segregation.
Subsequent work found that the protein products of these genes were required
for proteolytic destruction of speciic proteins, which triggers chromosome segregation.
8–127
As a irst step toward identiication of proteins that play a role in ATPdependent proteolytic destruction, cell extracts were passed over an
ion-exchange column. he ion-exchange column was then washed with
bufer and eluted with high salt. his resulted in two fractions: proteins
that bound to the column (bound fraction), and proteins that lowed
195
Chapter 8: Analyzing Cells, Molecules, and Systems
196
through the column (unbound fraction). No proteolytic activity could be
detected in either of these fractions, even though the starting extract had
robust activity. What should you do next?
A. Carry out additional puriication steps and test for activity.
B. Combine bound and unbound fractions and test for activity.
C. Improve the sensitivity of the assay used to detect proteolysis.
D. Use a gel-iltration column instead of an ion-exchange column.
E. Give up and go home.
Analysis of the unbound fraction identiied the small protein ubiquitin as
a key factor. he investigators then sought to discover additional proteins
in the cell lysates that work with ubiquitin in the proteolytic pathway.
Which one of the following techniques do you suppose they applied?
A. A BLAST search
B. Ainity chromatography
C. Gel-iltration chromatography
D. SDS gel electrophoresis
8–128
To isolate yeast mutants that cannot traverse the cell-division cycle, what
kind of mutation did the investigators need to obtain?
A. Conditional mutation
B. Gain-of-function mutation
C. Loss-of-function mutation
D. Null mutation
8–129
he Cdc16, Cdc23, and Cdc27 mutants displayed identical phenotypes,
which suggested that the encoded proteins work together to execute
common functions. Investigators hypothesized that they exist in a multiprotein complex. What technique would you use to test this hypothesis
most rapidly?
A. Co-immunoprecipitation
B. Ion-exchange chromatography
C. Nuclear magnetic resonance
D. X-ray difraction
8–130
Cdc16, Cdc23, and Cdc27 were found to be components of the anaphasepromoting complex (APC). he APC is an E3 complex that targets speciic
proteins for destruction during mitosis. Investigators hypothesized that
the APC triggers chromosome segregation by destroying a protein called
securin, which inhibits chromosome segregation. To test this hypothesis,
they arrested yeast cells in G1 phase to synchronize the cell population.
By releasing the cells from the G1 arrest, they could study a population
of cells going through the cell cycle in synchrony. Starting with the synchronized cells, which one of the following techniques would be most
useful for testing whether securin is proteolytically destroyed during the
cell cycle?
A. Hybridization
B. Immunoprecipitation
C. Mass spectrometry
D. Western blotting
8–131
Chapter 9
197
CHAPTER
9
Visualizing Cells
LOOKING AT CELLS IN THE LIGHT MICROSCOPE
TERMS TO LEARN
bright-field microscope
cell doctrine
confocal microscope
dark-field microscope
differential-interference-contrast
microscope
fluorescence microscope
fluorescence recovery after
photobleaching (FRAP)
fluorescence resonance energy
transfer (FRET)
green fluorescent protein (GFP)
image processing
ion-sensitive indicator
light microscope
limit of resolution
microelectrode
phase-contrast microscope
photoactivation
superresolution
DEFINITIONS
Match each deinition below with its term from the list above.
9–1
Fluorescent protein (from a jellyish) that is widely used as a marker for
monitoring the movements of proteins in living cells.
9–2
he minimal separation between two objects at which they appear distinct.
9–3
he normal light microscope in which the image is obtained by simple
transmission of light through the object being viewed.
9–4
Computer treatment of images gained from microscopy that reveals
information not immediately visible to the eye.
9–5
Similar to a light microscope but the illuminating light is passed through
one set of ilters before the specimen, to select those wavelengths that
excite the dye, and through another set of ilters before it reaches the eye,
to select only those wavelengths emitted when the dye luoresces.
9–6
Type of light microscope that produces a clear image of a given plane
within a solid object. It uses a laser beam as a pinpoint source of illumination and scans across the plane to produce a two-dimensional optical
section.
9–7
Technique for monitoring the closeness of two luorescently labeled
molecules (and thus their interaction) in cells.
9–8
Piece of ine glass tubing, pulled to an even iner tip, that is used to inject
electric current into cells.
IN THIS CHAPTER
LOOKING AT CELLS IN THE
LIGHT MICROSCOPE
LOOKING AT CELLS AND
MOLECULES IN THE ELECTRON
MICROSCOPE
198
Chapter 9: Visualizing Cells
TRUE/FALSE
G
Decide whether each of these statements is true or false, and then explain why.
9–9
F
Because the DNA double helix is only 2 nm wide—well below the limit of
resolution of the light microscope—it is impossible to see chromosomes
in living cells without special stains.
9–10
Monoclonal antibodies are superior to the mixture of antibodies in
standard antisera raised against the same protein because they are absolutely speciic for that protein.
9–11
Caged molecules can be introduced into a cell and then activated by a
strong pulse of laser light at the precise time and cellular location chosen
by the experimenter.
9–12
Superresolution techniques allow luorescently tagged molecules to be
imaged to accuracies an order of magnitude below the classic difraction
limit to resolution.
E
D
C
B
THOUGHT PROBLEMS
9–13
Examine the diagram of the light microscope in Figure 9–1. Identify and
label the eyepiece, the condenser, the light source, the objective, and the
specimen. At what two points in the light path is the image of the specimen magniied?
9–14
Why is it important to keep dust, ingerprints, and other smudges of the
lenses of a light microscope?
9–15
When light enters a medium with a diferent optical density, it bends in
a direction that depends on the refractive indices of the two media. Air
and glass, for example, have refractive indices of 1.00 and 1.51, respectively. When light enters glass—the medium with the higher refractive
index—it bends toward a line drawn normal to the surface (Figure 9–2A).
Conversely, when light exits glass into air, it bends away from the normal
line (Figure 9–2B). Using these principles, draw the paths of two parallel
light rays that pass through the hemispherical glass lens shown in Figure
9–2C. Will the two rays converge or diverge? Would the result be any different if the glass lens were lipped so that light entered the lat surface?
9–16
he diagrams in Figure 9–3 show the paths of light rays passing through
a specimen with a dry lens and with an oil-immersion lens. Ofer an
(A)
normal
(B)
A
Figure 9–1 Schematic diagram of a light
microscope (Problem 9–13).
normal
air
glass
glass
air
DRY LENS
OIL-IMMERSION LENS
(C)
glass
air
Figure 9–2 Refraction of light at air–glass
interfaces (Problem 9–15). (A) A ray of light
passing from air to glass. (B) A ray of light
passing from glass to air. (C) Two parallel rays
of light entering a glass lens.
air
objective
lens
cover slip
slide
oil
Figure 9–3 Paths of light rays through dry and oil-immersion lenses (Problem
9–16). The red circle at the origin of the light rays is the specimen.
LOOKING AT CELLS IN THE LIGHT MICROSCOPE
199
Figure 9–4 Diagram of the human eye
(Problem 9–17).
iris
vitreous
humor
retina
cornea
lens
aqueous
humor
explanation for why oil-immersion lenses should give better resolution.
Air, glass, and oil have refractive indices of 1.00, 1.51, and 1.51, respectively.
9–17
Figure 9–4 shows a diagram of the human eye. he refractive indices of
the components in the light path are: cornea 1.38, aqueous humor 1.33,
crystalline lens 1.41, and vitreous humor 1.38. Where does the main
refraction—the main focusing—occur? What role do you suppose the
lens plays?
9–18
Why do humans see so poorly under water? And why do goggles help?
9–19
Reading through a beaker illed with clear glass balls is impossible
(Figure 9–5). Do you suppose it would help to ill the beaker with water?
With immersion oil? Explain your reasoning.
9–20
Examine the four photomicrographs of the same cell in Figure 9–6.
Match each image to the technique listed below.
1. Bright-ield microscopy
2. Dark-ield microscopy
3. Nomarski diferential-interference-contrast microscopy
4. Phase-contrast microscopy
9–21
Explain the diference between resolution and magniication.
9–22
Many luorescent dyes that stain DNA require excitation by ultraviolet
light. Hoechst 33342, for example, binds to DNA and absorbs light at 352
nm and emits at 461 nm. If you want to use this membrane-permeant dye
in living cells, do you have to worry about ultraviolet light damage to the
DNA, which absorbs light maximally at around 260 nm? Why or why not?
(A)
(B)
(C)
(D)
Figure 9–5 Viewing an eye chart through
a beaker illed with clear glass balls
(Problem 9–19).
Figure 9–6 Four photomicrographs of the
same cell (Problem 9–20).
50 µm
200
Chapter 9: Visualizing Cells
(A)
barrier
filter
barrier
filter
light
source
beam-splitter
absorbance or emission
(B)
+
CH3NH
+
NH
N
1.5
+
NH
N
H
N
H
1.0
OCH2CH3
0.5
absorbance
0
280
320
360
emission
400 440 480 520
wavelength (nm)
560
600
Figure 9–7 Fluorescence microscopy (Problem 9–24). (A) The light path through a luorescence microscope. (B) Absorption
and luorescence emission spectra of Hoechst 33342 bound to DNA. The structure of Hoechst 33342 is shown above the
spectra.
9–23
Why is it, do you suppose, that a luorescent molecule, having absorbed
a single photon of light at one wavelength, always emits a photon at a
longer wavelength?
9–24
A luorescence microscope, which is shown schematically in Figure
9–7A, uses two ilters and a beam-splitting (dichroic) mirror to excite the
sample and capture the emitted luorescent light. Imagine that you wish
to view the luorescence of a sample that has been stained with Hoechst
33342, a common stain for DNA. he absorption and luorescence emission spectra for Hoechst 33342 are shown in Figure 9–7B.
A. Which of the following commercially available barrier ilters would you
select to place between the light source and the sample? Which one
would you place between the sample and the eyepiece?
1. A ilter that passes wavelengths between 300 nm and 380 nm.
2. A ilter that passes all wavelengths above 420 nm.
3. A ilter that passes wavelengths between 450 nm and 490 nm.
4. A ilter that passes all wavelengths above 515 nm.
5. A ilter that passes wavelengths between 510 nm and 560 nm.
6. A ilter that passes all wavelengths above 590 nm.
B. How would you design your beam-splitting mirror? Which wavelengths
would be relected and which would be transmitted?
9–25
Antibodies that bind to speciic proteins are important tools for deining the locations of molecules in cells. he sensitivity of the primary
antibody—the antibody that reacts with the target molecule—is often
enhanced by using labeled secondary antibodies that bind to it. What
are the advantages and disadvantages of using secondary antibodies that
carry luorescent tags versus those that carry bound enzymes?
9–26
he green luorescent protein (GFP) was isolated as a cDNA from a species of jellyish that glows green. When the cDNA for GFP was introduced
into bacteria, the colonies they formed glowed pale green under ultraviolet light. In these early studies, the following pertinent observations
provided important insights into how GFP becomes luorescent.
1. When bacteria are grown anaerobically, they express normal amounts of
GFP but it is not luorescent.
2. he denatured GFP found in insoluble protein aggregates (inclusion
bodies) in bacteria is not luorescent.
3. he rate of appearance of luorescence follows irst-order kinetics, with a
time constant that is independent of the concentration of GFP.
LOOKING AT CELLS IN THE LIGHT MICROSCOPE
201
Figure 9–8 A rainbow of colors produced
by modiied GFPs (Problem 9–27).
4. Random mutations introduced into the cDNA coding for GFP produced
some proteins with appreciably brighter luorescence and some with different colors.
Comment on what each of these observations says about GFP luorescence.
9–27
Figure 9–8 shows a series of modiied GFPs that emit light in a range of
colors. How do you suppose the exact same chromophore can luoresce
at so many diferent wavelengths?
CALCULATIONS
9–28
he resolving power of a light microscope depends on the width of the
cone of light that illuminates the specimen, the wavelength of the light
used, and the refractive index of the medium separating the specimen
from the objective and condenser lenses, according to the formula
resolution = 0.61 λ
n sin θ
where λ equals the wavelength of light used, n is the refractive index, and
θ is half the angular width of the cone of rays collected by the objective
lens. Assuming an angular width of 120° (θ = 60°), calculate the various
resolutions you would expect if the sample were illuminated with violet
light (λ = 0.4 μm) or red light (λ = 0.7 μm) in a refractive medium of air
(n = 1.00) or oil (n = 1.51). Which of these conditions would give you the
best resolution?
9–29
At a certain critical angle, which depends on the refractive indices of
the two media, light from the optically denser medium will be bent suficiently that it cannot escape and will be relected back into the denser
medium. he formula that describes refraction is
ni sin θi = nt sin θt
where ni and nt are the refractive indices of the incident and transmitting media, respectively, and θi and θt are the incident and transmitted
angles, respectively, as shown in Figure 9–9A. For a glass–air interface,
calculate the incident angle for a light ray that is bent so that it is parallel
to the surface (Figure 9–9B). he refractive indices for air and glass are
1.00 and 1.51, respectively.
(A)
(B)
θt
nt
θt
air
glass
ni
θi
θi
Figure 9–9 Refraction at the interface
between media with different refractive
indices (Problem 9–29). (A) The
relationship between incident and
transmitted angles for materials with
different refractive indices such that
ni > nt. (B) The incident angle for a
glass–air interface such that the
transmitted (bent) ray is parallel to
the interface.
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Chapter 9: Visualizing Cells
TABLE 9–1 Excitation properties of a few common biological
fluorophores (Problem 9–30).
opea absorption
maximum (nm)
Tpeb absorption
maximum (nm)
Emission
maximum (nm)
DAPI
358
685
461
ER-Tracker
374
728
575
Mito-Tracker
579
1133
599
Alexa Fluor 488
491
985
515
FITC-IgG
490
947
525
Fluorophore
aOne-photon excitation; bTwo-photon excitation.
DATA HANDLING
9–30
Fluorescent molecules can be excited by a single high-energy photon or
by multiple lower-energy photons. A list of commonly used biological
luorophores is given in Table 9–1. Why do you suppose that the absorption maximum for two-photon excitation is about twice that for one-photon excitation?
9–31
You wish to attach luorescent tags to two diferent proteins so that you
can follow them independently. he excitation and emission spectra for
cyan, green, and yellow luorescent proteins (CFP, GFP, and YFP) are
shown in Figure 9–10. Can you pick any pair of these proteins? Or are
some pairs better than others? Explain your answer.
9–32
Consider a luorescent detector designed to report the cellular location
of active protein tyrosine kinases. A blue (cyan) luorescent protein (CFP)
and a yellow luorescent protein (YFP) were fused to either end of a hybrid
CFP
GFP YFP
percent of maximum
EXCITATION
EMISSION
GFP YFP
percent of maximum
CFP
400
450
500
550
wavelength (nm)
600
Figure 9–10 Excitation and emission
spectra for CFP, GFP, and yFP
(Problem 9–31).
LOOKING AT CELLS IN THE LIGHT MICROSCOPE
(A) REPORTER
203
(B) FRET
434 nm
476 nm
+ phosphatase
1.3
CF
Abl + ATP
YFP/CFP
P
substrate
peptide
1.2
1.1
omit Abl or ATP
YFP
phosphotyrosinebinding protein
1.0
0
5
10
15
20
25
30
Figure 9–11 Fluorescent reporter
protein designed to detect tyrosine
phosphorylation (Problem 9–32).
(A) Domain structure of reporter protein.
Four domains are indicated: CFP, yFP,
tyrosine kinase substrate peptide, and
a phosphotyrosine-binding domain.
(B) FRET assay. yFP/CFP is normalized
to 1.0 at time zero. The reporter was
incubated in the presence (or absence)
of Abl and ATP for the indicated times.
Arrow indicates time of addition of a
tyrosine phosphatase.
time (hours)
protein domain. he hybrid protein segment consisted of a substrate
peptide recognized by the Abl protein tyrosine kinase and a phosphotyrosine-binding domain (Figure 9–11A). Stimulation of the CFP domain
does not cause emission by the YFP domain when the domains are separated. When the CFP and YFP domains are brought close together, luorescence resonance energy transfer (FRET) allows excitation of CFP to
stimulate emission by YFP. FRET shows up experimentally as an increase
in the ratio of emission at 526 nm versus 476 nm (YFP/CFP) when CFP is
excited by 434 nm light.
Incubation of the reporter protein with Abl protein tyrosine kinase
in the presence of ATP gave an increase in YFP/CFP emission (Figure
9–11B). In the absence of ATP or the Abl protein, no FRET occurred.
FRET was also eliminated by addition of a tyrosine phosphatase (Figure
9–11B). Describe as best you can how the reporter protein detects active
Abl protein tyrosine kinase.
9–33
Cells activate Abl protein tyrosine kinase in response to platelet-derived
growth factor (PDGF). PDGF binds to the PDGF receptor, which activates
Src, which then activates Abl. It is unclear where in the cell Abl is active,
but one of the consequences of PDGF stimulation is the appearance of
membrane rules. To investigate this question, the reporter construct
described in Problem 9–32 was transfected into cells, which were then
stimulated by addition of PDGF. Using luorescence microscopy, YFP
emission in response to CFP excitation (FRET) was followed in diferent
parts of the cell, with the results shown in Figure 9–12. What can you
infer about the cellular distribution of active Abl protein tyrosine kinase
in response to PDGF?
9–34
You are using a cameleon indicator to measure intracellular concentrations of Ca2+. he indicator is composed of a central calmodulin domain,
which converts from an extended form to a much more compact form
upon calcium binding, and two lanking luorescent proteins, with CFP
2.3
ruffles
YFP/CFP
2.2
cytoplasm
2.1
2.0
1.9
nucleus
1.8
0
10
20
30
40
50
time after adding PDGF (minutes)
60
Figure 9–12 Time course of FRET in
various parts of the cell after addition
of PDGF (Problem 9–33). FRET was
measured as the increase in the ratio of
emission at 526 nm to that at 476 nm
(yFP/CFP) when CFP was excited by
434 nm light.
204
Chapter 9: Visualizing Cells
Time (milliseconds)
0
300
600
900
1200
1500
attached at one end and YFP attached at the other. You have expressed
this indicator in cells and now wish to measure intracellular changes in
Ca2+ concentration in response to the biological process you are studying. he instructions say to excite the cameleon at 440 nm and to measure
emission at 535 nm. How do you suppose the cameleon indicator works?
9–35
he formation of clathrin-coated vesicles is initiated by the assembly of a
basketlike framework of clathrin on the underside of the cell membrane,
which distorts the membrane into a shallow pit. hese clathrin-coated
pits ultimately pinch of to form clathrin-coated vesicles. You want to
understand how pits are converted into vesicles, and have engineered a
cell line that expresses clathrin–GFP. To focus on events at the membrane
surface, you use total internal relection luorescence (TIRF) microscopy
to examine the cell membrane and a thin slice of adjacent cytosol. You
are excited to ind that the samples show many green dots under the
microscope and that some dots disappear over time (Figure 9–13).
A. Why do you suppose some dots remain unchanged, while others disappear?
B. You are concerned that the green dots may not be functional assemblies,
but some kind of artifact. A colleague suggests that you test whether the
green dots really represent clathrin-coated structures by incubating the
cells with transferrin that is tagged with a red luorophore. Transferrin,
which carries iron ions, binds to its receptors on the cell surface and is
taken into the cell in clathrin-coated vesicles. What might you expect to
see if you carried out this experiment?
LOOKING AT CELLS AND MOLECULES IN THE
ELECTRON MICROSCOPE
TERMS TO LEARN
cryoelectron microscopy
electron microscope (EM)
electron-microscope tomography
immunogold electron microscopy
negative staining
scanning electron microscope (SEM)
single-particle reconstruction
DEFINITIONS
Match each deinition below with its term from the list above.
9–36
A contrast-enhancing technique for the electron microscope in which
a heavy-metal salt is used to create a reverse, or negative, image of the
object.
9–37
Type of microscope that uses a beam of electrons to create an image.
9–38
Electron microscopy technique in which the objects to be viewed, such
as macromolecules and viruses, are rapidly frozen.
9–39
Type of electron microscope that produces an image of the surface of an
object.
Figure 9–13 TIRF microscopy of cells
expressing clathrin–GFP (Problem 9–35).
Each frame in this time course represents
300 msec. The yellow arrow indicates a
dot that disappears during the 1.5 sec
observation period. Green dots appear
white in this image.
LOOKING AT CELLS AND MOLECULES IN THE ELECTRON MICROSCOPE
9–40
205
Electron microscopy technique in which cellular structures or molecules
of interest are labeled with antibodies tagged with electron-dense gold
particles, which show up as black spots on the image.
TRUE/FALSE
Decide whether this statement is true or false, and then explain why.
9–41
Transmission electron microscopy (TEM) and scanning electron microscopy (SEM) can both be used to examine a structure in the interior of a
thin section; TEM provides a projection view, while SEM captures electrons scattered from the structure and gives a more three-dimensional
view.
THOUGHT PROBLEMS
9–42
A major challenge to electron microscopists from the beginning was to
convince others that what they observed in micrographs truly relected
structures that were originally present in the living cell. Outline a current
approach to this problem.
9–43
he technique of negative staining uses heavy metals such as uranium to
provide contrast. If the heavy metals don’t actually bind to deined biological structures (which they don’t), how is it that they can help to make
such structures visible?
9–44
Heavy metals can be used to highlight the surface features of a sample.
In the technique called metal shadowing, a heavy metal such as platinum is evaporated at a shallow angle onto the specimen so as to deposit
a thin ilm. he thickness of the ilm depends on the surface features of
the sample and on their orientation relative to the source of platinum. By
stabilizing the metal ilm and eliminating the original specimen, it is possible to image the metal replica of the sample by transmission electron
microscopy, as shown in the micrographs in Figure 9–14.
It is sometimes diicult to tell bumps from pits in such micrographs
just by looking at the pattern of shadows, as illustrated in Figure 9–14 for
a set of shaded circles. In Figure 9–14A, the circles appear to be bumps;
however, when the picture is simply turned upside down (Figure 9–14B),
the circles seem to be pits. his is a classic illusion. he same illusion is
present in metal shadowing, as shown in Figure 9–14C and D. In one
micrograph, the membrane appears to be covered in bumps, while in the
same micrograph, turned upside down, the membrane looks heavily pitted. Is it possible for an electron microscopist to be sure that one view is
correct, or is it all arbitrary? Explain your reasoning.
(A)
(B)
(C)
(D)
Figure 9–14 Bumps and pits (Problem 9–44). (A) Shaded circles that look like bumps. (B) Shaded circles that look like
pits. (C) An electron micrograph oriented so that it appears to be covered with bumps. (D) The same electron micrograph
oriented so that it appears to be covered with pits.
206
Chapter 9: Visualizing Cells
9–45
Nuclear pore complexes, which are large assemblages of more than 30
diferent proteins, mediate the exchange of macromolecules between
the nucleus and cytoplasm. You have gently isolated nuclei from Dictyostelium discoideum and frozen them in vitreous ice for examination
by cryoelectron tomography. You obtain a tomogram of a nucleus and
combine the images from 267 nuclear pore complexes. You expect that
individual nuclei will have been arrested in diferent states of transport
because the isolated nuclei were shown to be competent for transport.
Assuming that parts of the structure lex and move during the transport
process, how do you suppose that averaging structures in diferent states
will afect your inal picture? Can you think of a way to improve the quality of the image you would get from this data set?
CALCULATIONS
9–46
he practical resolving power of modern electron microscopes is around
0.1 nm. he major reason for this constraint is the small numerical aperture (n sin θ), which is limited by θ (half the angular width of rays collected at the objective lens). Assuming that the wavelength (λ) of the
electron is 0.004 nm and that the refractive index (n) is 1.0, calculate the
value for θ. How does that value compare with a θ of 60°, which is typical
for light microscopes?
resolution = 0.61 λ
n sin θ
DATA HANDLING
9–47
You are studying two proteins that you think may be components of gap
junctions, which are structures in membranes that allow adjacent cells to
exchange small molecules. When cells are very rapidly frozen and then
cracked with a knife blade, the cells often fracture along membrane surfaces. When the freeze-fractured cells are viewed in the electron microscope, gap junctions show up clearly as specialized areas densely populated with membrane particles. To decide whether your proteins are
components of gap junctions, you have prepared antibodies against each
of them. To one antibody you’ve attached 15 nm gold particles; to the
other, 10 nm gold particles. You prepare freeze-fractured cells for electron microscopy and then incubate them with your gold-tagged antibodies, with the results shown in Figure 9–15. Do these results indicate that
both proteins are part of gap junctions? Why or why not?
200 nm
Figure 9–15 A freeze-fracture electron
micrograph of a gap junction (Problem
9–47). The central densely packed area
outlined in white is the gap junction. The
two lealets of the plasma membrane are
indicated as EF (for external face) and
PF (for protoplasmic face). Gold particles
show up as black dots.
LOOKING AT CELLS AND MOLECULES IN THE ELECTRON MICROSCOPE
9–48
207
Aquaporin water channels, which are known to be located in the plasma
membrane, play a major role in water metabolism and osmoregulation
in many cells, including nerve cells of the brain and spinal cord. To determine their structural organization in the membrane, you have prepared a
highly speciic antibody against the protein AQP4, which forms the water
channels in the astrocytes of the brain. You prepare a freeze-fractured
sample from the brain, incubate it with gold-tagged antibodies against
AQP4, and examine it by electron microscopy (Figure 9–16).
A. Are the gold particles (black dots) consistently associated with any particular structure?
B. Are there any examples of black dots that are not associated with these
structures? Are there any examples of structures that do not have black
dots? How do your answers to these questions afect your conidence
that the structure you’ve identiied is the aquaporin water channel?
MCAT STYLE
Passage (Questions 9–49 to 9–51)
Many key advances in cell biology were dependent upon technical advances in
microscopy. his was particularly true in the case of the cytoskeleton, the dynamic
network of cytoplasmic ibers that is responsible for cell motility and chromosome
movement during mitosis. he two key components of the cytoskeleton—actin
ilaments and microtubules—assemble by polymerization of individual subunits.
Actin ilaments and microtubules were irst seen by electron microscopy, which
showed that they have diameters of 7 nm and 25 nm, respectively, and can form
very long polymers. Later development of light microscopy techniques for imaging individual microtubules and actin ilaments under natural conditions revolutionized the ield.
9–49
Imagine that you are developing techniques to visualize individual actin
ilaments by light microscopy for the irst time. Which of the following
approaches would you choose?
I. Assemble actin ilaments from luorescently tagged subunits and image
them with a luorescence microscope.
II. Visualize actin ilaments by using diferential-interference-contrast
microscopy.
III. Combine standard light microscopic techniques with digital image processing to enhance resolution of actin ilaments.
A. I
B. II
C. III
D. I, II, and III
9–50
What would be the apparent diameter of actin ilaments imaged using
the approach or approaches you selected in the previous question?
A. 7 nm
B. 50 nm
C. 100 nm
D. 200 nm
9–51
Microscopy has also played a key role in answering mechanistic questions about how enzymes work. Consider, for example, ATP synthase, the
remarkable multisubunit enzyme responsible for ATP synthesis in the
inner mitochondrial membrane. he structure of ATP synthase revealed
a central stalk surrounded by other subunits that form a ring 10 nm in
diameter. It was hypothesized that rotation of the stalk relative to the ring
subunits—powered by proton low across the inner membrane—generated suicient mechanical energy to force the conformational changes
needed to drive ATP synthesis. You would like to test this hypothesis by
100 nm
Figure 9–16 Freeze-fracture micrograph
of an astrocyte membrane labeled with
gold-tagged antibodies against AQP4
(Problem 9–48).
208
Chapter 9: Visualizing Cells
directly visualizing rotation of the stalk relative to the ring. Which one of
the following ideas do you think would be the best approach for detecting rotation of the stalk?
A. Attach a 2 μm-long actin ilament assembled from luorescent subunits
to the stalk and view its rotation by luorescence microscopy.
B. Attach green luorescent protein (GFP) to the proteins in the ring and use
luorescence microscopy to observe rotation of the ring around the stalk.
C. Attach GFP to the stalk and RFP (red luorescent protein) to the ring and
observe their relative rotation by luorescence microscopy.
D. Attach multiple luorescently labeled antibodies to the stalk and then
observe rotation of the stalk by luorescence microscopy.
Chapter 10
209
CHAPTER
Membrane Structure
10
THE LIPID BILAYER
IN THIS CHAPTER
TERMS TO LEARN
THE LIPID BILAYER
amphiphilic
cholesterol
ganglioside
glycolipid
hydrophilic
hydrophobic
lipid bilayer
lipid droplet
lipid raft
liposome
phosphoglyceride
phospholipid
plasma membrane
DEFINITIONS
Match the deinition below with its term from the list above.
10–1
Artiicial phospholipid bilayer vesicle formed from an aqueous suspension of phospholipid molecules.
10–2
Small region of the plasma membrane enriched in sphingolipids and
cholesterol.
10–3
Any glycolipid having one or more sialic acid residues in its structure;
especially abundant in the plasma membranes of nerve cells.
10–4
Having both hydrophobic and hydrophilic regions, as in a phospholipid
or a detergent molecule.
10–5
he main type of phospholipid in animal cell membranes, with two fatty
acids and a polar head group attached to a three-carbon glycerol backbone.
10–6
Lipid molecule with a characteristic four-ring steroid structure that is an
important component of the plasma membranes of animal cells.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
10–7
Although lipid molecules are free to difuse in the plane of the bilayer,
they cannot lip-lop across the bilayer unless enzyme catalysts called
phospholipid translocators are present in the membrane.
10–8
All of the common phospholipids—phosphatidylcholine, phosphatidylethanolamine, phosphatidylserine, and sphingomyelin—carry a positively charged moiety on their head group, but none carry a net positive
charge.
10–9
Glycolipids are never found on the cytoplasmic face of membranes in
living cells.
MEMBRANE PROTEINS
Chapter 10: Membrane Structure
210
C
O
H
H
CH3
H 3C
Figure 10–1 Icelike cage of water
molecules around a hydrophobic solute
(Problem 10–10).
H
H
CH3
H
H
O
H
O
H
2-methyl propane
H
O
O
H
δ
H
O
_
H
O
H
H
δ+
H
H
O
O
H
δ+
water
H
H
2-methyl propane in water
THOUGHT PROBLEMS
10–10
Hydrophobic solutes are said to “force the adjacent water molecules to
reorganize into icelike cages” (Figure 10–1). his statement seems paradoxical because water molecules do not interact with hydrophobic solutes. How could water molecules “know” about the presence of a hydrophobic solute and change their behavior to interact diferently with one
another? Discuss this seeming paradox and develop a clear concept of
what is meant by an “icelike” cage. How does it compare to ice? Why
would such a cagelike structure be energetically unfavorable relative to
pure water?
10–11
When a lipid bilayer is torn, why does it not seal itself by forming a “hemimicelle” cap at the edges, as shown in Figure 10–2?
10–12
Five students in your class always sit together in the front row. his could
be because (1) they really like each other or (2) nobody else in your class
wants to sit next to them. Which explanation holds for the assembly of a
lipid bilayer? Explain your answer. If the lipid bilayer assembled for the
opposite reason, how would its properties difer?
10–13
Predict the properties of a lipid bilayer in which all of the hydrocarbon
chains were saturated. What would be the properties if all of the hydrocarbon chains were unsaturated?
10–14
What is meant by the term “two-dimensional luid”?
10–15
Margarine is made from vegetable oil by a chemical process. Do you suppose this process converts saturated fatty acids to unsaturated ones, or
vice versa? Explain your answer.
Which one of the phospholipids listed below is present in very small
quantities in the plasma membranes of mammalian cells?
A. Phosphatidylcholine
B. Phosphatidylethanolamine
C. Phosphatidylinositol
D. Phosphatidylserine
E. Sphingomyelin
10–16
tear in bilayer
seal with hemi-micelle cap
Figure 10–2 A torn lipid bilayer sealed
with a hypothetical “hemi-micelle” cap
(Problem 10–11).
THE LIPID BILAYER
10–17
211
Many snake venoms contain enzymes that cause red blood cells to lyse.
Imagine that you have puriied such an enzyme. When you add the puriied enzyme to red blood cells, you ind that in addition to cell lysis,
choline with a phosphate group attached to it is released, as well as diacylglycerol (glycerol with two fatty acid chains attached). What molecule
is cleaved by the enzyme to cause cell lysis?
Predict which of the following organisms will have the highest percentage of unsaturated fatty acid chains in their membranes. Explain your
answer.
A. Antarctic ish
B. Desert iguana
C. Human being
D. Polar bear
E. hermophilic bacterium
10–18
10–19
If lipid rafts form because membrane components such as sphingolipids
and cholesterol molecules preferentially associate with one another, why
do you think it is that they aggregate into multiple tiny rafts instead of into
a single large one?
10–20
he lipid bilayers found in cells are luid, yet asymmetrical in the composition of the monolayers. Is this a paradox? Explain your answer.
10–21
Phosphatidylserine, which is normally conined to the cytoplasmic monolayer of the plasma membrane lipid bilayer, is redistributed to the outer
monolayer during apoptosis. How is this redistribution accomplished?
CALCULATIONS
10–22
If a lipid raft is typically 70 nm in diameter and each lipid molecule has
a diameter of 0.5 nm, about how many lipid molecules would there be
in a lipid raft composed entirely of lipid? At a ratio of 50 lipid molecules
per protein molecule (50% protein by mass), how many proteins would
be in a typical raft? (Neglect the loss of lipid from the raft that would be
required to accommodate the protein.)
O
H
N
H
H
H
DATA HANDLING
nitroxide
radical
A classic paper studied the behavior of lipids in the two monolayers of
a membrane by labeling individual molecules with nitroxide groups,
which are stable free radicals (Figure 10–3). hese spin-labeled lipids
can be detected by electron spin resonance (ESR) spectroscopy, a technique that does not harm living cells. Spin-labeled lipids are introduced
into small lipid vesicles, which are then fused with cells, thereby transferring the labeled lipids into the plasma membrane.
H
N
H
H
H
phospholipid 2
N
O
H
phospholipid 1
H
he two spin-labeled phospholipids shown in Figure 10–3 were incorporated into intact human red blood cell membranes in this way. To determine whether they were introduced equally into the two monolayers of
the bilayer, ascorbic acid (vitamin C), which is a water-soluble reducing agent that does not cross membranes, was added to the medium to
destroy any nitroxide radicals exposed on the outside of the cell. he ESR
signal was followed as a function of time in the presence and absence of
ascorbic acid as indicated in Figure 10–4A and B.
A. Ignoring for the moment the diference in extent of loss of ESR signal,
ofer an explanation for why phospholipid 1 (Figure 10–4A) reacts faster
with ascorbate than does phospholipid 2 (Figure 10–4B). Note that phospholipid 1 reaches a plateau in about 15 minutes, whereas phospholipid
2 takes almost an hour.
O
H
H
10–23
Figure 10–3 Structures of two nitroxidelabeled lipids (Problem 10–23). The
nitroxide radical is shown at the top,
and its position of attachment to the
phospholipids is shown below.
Chapter 10: Membrane Structure
212
(A) PHOSPHOLIPID 1 – RED CELLS
(B) PHOSPHOLIPID 2 – RED CELLS
O
H
H
100
– ascorbate
100
75
75
− ascorbate
50
+ ascorbate
50
25
+ ascorbate
0
0
10
20
30
25
0
0
− ascorbate
100
2
3
− ascorbate
100
75
75
+ ascorbate
+ ascorbate
50
50
25
25
0
0
1
(D) PHOSPHOLIPID 2 – GHOSTS
(C) PHOSPHOLIPID 1 – GHOSTS
signal intensity (%)
N
H
signal intensity (%)
O
H
N
H
H
H
H
10
20
time (minutes)
30
0
0
1
2
time (hours)
3
B. To investigate the diference in extent of loss of ESR signal with the two
phospholipids, the experiments were repeated using red cell ghosts that
had been resealed to make them impermeable to ascorbate (Figure
10–4C and D). Resealed red cell ghosts are missing all of their cytoplasm,
but have an intact plasma membrane. In these experiments, the loss of
ESR signal for both phospholipids was negligible in the absence of ascorbate and reached a plateau at 50% in the presence of ascorbate. What
do you suppose might account for the diference in extent of loss of ESR
signal in experiments with red cell ghosts (Figure 10–4C and D) versus
those with normal red cells (Figure 10–4A and B).
C. Were the spin-labeled phospholipids introduced equally into the two
monolayers of the red cell membrane?
Fluorescence resonance energy transfer (FRET) has been used to investigate the existence of lipid rafts in living cells. To test for the presence
of lipid rafts by FRET, you use two diferent cell lines: one that expresses
a glycosylphosphatidylinositol (GPI)-anchored form of the folate receptor, and one that expresses the transmembrane-anchored form. Folate
receptors can be made luorescent by addition of a luorescent folate
analog. Cells tagged in this way show variation in luorescence intensity
over their surface because of chance variations in the density of labeled
receptors. his allows diferent densities of receptors to be analyzed by
examining diferent places in the same cell. he proximity of labeled
receptors can be determined by FRET, which depends on the distance
between receptors. he ratio of FRET to direct luorescence gives diferent expectations for dispersed receptors versus receptors that are clustered together, as depicted in Figure 10–5.
A. Explain the basis for the diference between the graphs of these expectations.
10–24
Figure 10–4 Decrease in ESR signal
intensity as a function of time in red cells
and red cell ghosts in the presence and
absence of ascorbate (Problem 10–23).
(A and B) Phospholipid 1 and
phospholipid 2 in red cells. (C and D)
Phospholipid 1 and phospholipid 2 in
red cell ghosts.
THE LIPID BILAYER
213
(A) RANDOM DISTRIBUTION
low density
density
high density
FRET/total
(B) NONRANDOM DISTRIBUTION
low density
density
high density
FRET/total
Figure 10–5 Expectations for FRET between tagged folate receptors at different
densities of receptors (Problem 10–24). (A) Randomly distributed receptors. The
red dots in the box represent luorescent receptors. (B) receptors clustered in
microdomains. Circles represent microdomains such as lipid rafts. Red dots
represent luorescent receptors.
B. he actual experiments showed that transmembrane-anchored folate
receptors followed the expectations shown in Figure 10–5A, whereas
the GPI-anchored folate receptors followed those in Figure 10–5B. Do
these experiments provide evidence for the existence of lipid rafts in the
plasma membrane? Why or why not?
he asymmetric distribution of phospholipids in the two monolayers
of the plasma membrane implies that very little spontaneous lip-lop
occurs or, alternatively, that any spontaneous lip-lop is rapidly corrected by phospholipid translocators that return phospholipids to their
appropriate monolayer. he rate of phospholipid lip-lop in the plasma
membrane of intact red blood cells has been measured to decide between
these alternatives.
One experimental measurement used the same two spin-labeled phospholipids described in Problem 10–23 (see Figure 10–3). To measure the
rate of lip-lop from the cytoplasmic monolayer to the outer monolayer,
red cells with spin-labeled phospholipids exclusively in the cytoplasmic
monolayer were incubated for various times in the presence of ascorbate
and the loss of ESR signal was followed. To measure the rate of lip-lop
from the outer to the cytoplasmic monolayer, red cells with spin-labeled
phospholipids exclusively in the outer monolayer were incubated for
various times in the absence of ascorbate and the loss of ESR signal was
followed. he results of these experiments are illustrated in Figure 10–6.
A. From the results in Figure 10–6, estimate the rate of lip-lop from the
cytoplasmic to the outer monolayer, and from the outer to the cytoplasmic monolayer. A convenient way to express such rates is as the half-time
of lip-lop—that is, the time it takes for half the phospholipids to lip-lop
from one monolayer to the other.
B. From what you learned about the behavior of the two spin-labeled phospholipids in Problem 10–23, deduce which one was used to label the
cytoplasmic monolayer of the intact red blood cells, and which one was
used to label the outer monolayer.
C. Using the information in this problem, propose a method to generate
intact red cells that contain spin-labeled phospholipids exclusively in
the cytoplasmic monolayer, and a method to generate cells spin-labeled
exclusively in the outer monolayer.
signal intensity (%)
10–25
outside
100
inside
50
10
0
1
2
4
5
3
time (hours)
6
7
Figure 10–6 Decrease in esr signal
intensity of red blood cells containing
spin-labeled phospholipids in the outer
monolayer (outside) and cytoplasmic
monolayer (inside) of the plasma
membrane (Problem 10–25).
214
Chapter 10: Membrane Structure
MEMBRANE PROTEINS
TERMS TO LEARN
bacteriorhodopsin
carbohydrate layer
cortex
detergent
glycosylphosphatidylinositol
(GPI) anchor
lectin
lumen
membrane-associated protein
membrane-bending protein
membrane protein
multipass transmembrane protein
single-pass transmembrane protein
spectrin
transmembrane protein
DEFINITIONS
Match the deinition below with its term from the list above.
10–26
Protein that binds tightly to a speciic sugar.
10–27
he outer coat of a eukaryotic cell, composed of oligosaccharides linked
to intrinsic plasma membrane glycoproteins and glycolipids, as well as
proteins that have been secreted and reabsorbed onto the cell surface.
10–28
Abundant protein associated with the cytosolic side of the plasma membrane in red blood cells, forming a rigid network that supports the membrane.
10–29
Protein whose polypeptide chain crosses the lipid bilayer more than
once.
10–30
Pigmented protein found in the plasma membrane of Halobacterium
halobium, where it pumps protons out of the cell in response to light.
10–31
he complicated cytoskeletal network in the cytosol just beneath the
plasma membrane.
10–32
Type of lipid linkage, formed as proteins pass through the endoplasmic
reticulum, by which some proteins are attached to the noncytosolic surface of the membrane.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
10–33
Whereas all the carbohydrate in the plasma membrane faces outward on
the external surface of the cell, all the carbohydrate on internal membranes faces toward the cytosol.
10–34
Human red blood cells contain no internal membranes other than the
nuclear membrane.
10–35
Although membrane domains with diferent protein compositions are
well known, there are at present no examples of membrane domains that
difer in lipid composition.
THOUGHT PROBLEMS
10–36
Which of the arrangements of membrane-associated proteins indicated
in Figure 10–7 have been found in biological membranes?
10–37
Which one of the following statements correctly describes the mass ratio
of lipids to proteins in membranes?
MEMBRANE PROTEINS
215
3
6
NH2
COOH
P
P
lipid
bilayer
CYTOSOL
7
1
COOH
2
5
4
8
Figure 10–7 A variety of possible
associations of proteins with a membrane
(Problem 10–36).
A. he mass of lipids greatly exceeds the mass of proteins.
B. he mass of proteins greatly exceeds the mass of lipids.
C. he masses of lipids and proteins are about equal.
D. he mass ratio of lipids to proteins varies widely in diferent membranes.
10–38
Name the three general types of lipid anchors that are used to attach proteins to membranes.
10–39
Monomeric single-pass transmembrane proteins span a membrane with
a single α helix that has characteristic chemical properties in the region
of the bilayer. Which of the three 20-amino-acid sequences listed below
is the most likely candidate for such a transmembrane segment? Explain
the reasons for your choice. (See Table 8, page 966, for one-letter amino
acid code; FAMILY VW is a convenient mnemonic for hydrophobic
amino acids.)
A. I T L I Y F G V M A G V I G T I L L I S
B. I T P I Y F G P M A G V I G T P L L I S
C. I T E I Y F G R M A G V I G T D L L I S
10–40
10–41
Consider a transmembrane protein complex that forms a hydrophilic
pore across the plasma membrane of a eukaryotic cell. he pore is made
of ive similar protein subunits, each of which contributes a membranespanning α helix to form the pore. Each α helix has hydrophilic amino
acid side chains on one side of the helix and hydrophobic amino acid
side chains on the opposite side. Propose a possible arrangement of
these ive α helices in the membrane.
Why is it that membrane-spanning protein segments are almost always α
helices or β barrels, but never disordered chains?
10–42
You are studying the binding of proteins to the cytoplasmic face of cultured neuroblastoma cells and have found a method that gives a good
yield of inside-out vesicles from the plasma membrane. Unfortunately,
your preparations are contaminated with variable amounts of right-sideout vesicles. Nothing you have tried avoids this problem. A friend suggests that you pass your vesicles over an ainity column made of lectin
coupled to solid beads. What is the point of your friend’s suggestion?
10–43
Detergents are small amphiphilic molecules—one end hydrophobic
and the other hydrophilic—that tend to form micelles in water. Examine
the structures of sodium dodecyl sulfate (SDS) and Triton X-100 in
Figure 10–8 and explain why the black portions are hydrophilic and the
blue sections are hydrophobic.
10–44
Why does a red blood cell membrane need proteins?
CH3
CH3
CH3
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
O
O S O
O– Na+
sodium (Na+)
dodecyl
sulfate (SDS)
C CH3
CH2
CH3
C CH3
H
H
H
H
O
CH2
CH2
O
CH2
CH2
O
9–10
CH2
CH2
OH
Triton X-100
Figure 10–8 The structures of SDS and
Triton X-100 (Problem 10–43).
Chapter 10: Membrane Structure
216
(A)
(B)
(C)
(D)
CYTOSOL
EXTRACELLULAR
SPACE
protrusion
10–45
Glycophorin, a protein in the plasma membrane of the red blood cell,
normally exists as a homodimer that is held together entirely by interactions between its transmembrane domains. Since transmembrane
domains are hydrophobic, how is it that they can associate with one
another so speciically?
10–46
Describe the diferent methods that cells use to restrict proteins to
speciic regions of the plasma membrane. Is a membrane with many
anchored proteins still luid?
10–47
hree mechanisms by which membrane-binding proteins bend a
membrane are illustrated in Figure 10–9A–C. As shown, each of these
cytosolic membrane-bending proteins would induce an invagination of
the plasma membrane. Could similar kinds of cytosolic proteins induce
a protrusion of the plasma membrane (Figure 10–9D)? Which ones?
Explain how they might work.
CALCULATIONS
10–48
In the membrane of a human red blood cell, the ratio of the mass of protein (average molecular weight 50,000) to phospholipid (average molecular weight 800) to cholesterol (molecular weight 386) is about 2:1:1.
How many lipid molecules (phospholipid + cholesterol) are there for
every protein molecule?
Estimates of the number of membrane-associated proteins per cell and
the fraction of the plasma membrane occupied by such proteins provide
a useful quantitative basis for understanding the structure of the plasma
membrane. hese calculations are straightforward for proteins in the
plasma membrane of a red blood cell because red cells are readily isolated from blood and they contain no internal membranes to confuse
the issue. Plasma membranes are prepared, the membrane-associated
proteins are separated by SDS polyacrylamide-gel electrophoresis, and
then they are stained with a dye (Coomassie Blue). Because the intensity
of color is roughly proportional to the mass of protein present in a band,
quantitative estimates can be made as shown in Table 10–1.
A. From the information in Table 10–1, calculate the number of molecules
of spectrin, anion exchanger (AE1), and glycophorin in an individual red
blood cell. Assume that 1 mL of red cell ghosts contains 1010 cells and 5
mg of total membrane protein.
10–49
TABLE 10–1 Proportion of stain associated with three membrane-associated
proteins (Problem 10–49).
Protein
Molecular weight
Percent of stain
Spectrin
250,000
25
AE1
100,000
30
Glycophorin
30,000
2.3
Figure 10–9 Bending of the plasma
membrane by cytosolic proteins (Problem
10–47). (A) Insertion of a protein “inger”
into the cytosolic lealet of the membrane.
(B) Binding of lipids to the curved
surface of a membrane-binding protein.
(C) Binding of membrane proteins to
membrane lipids with large head groups.
(D) a segment of the plasma membrane
showing a protrusion.
MEMBRANE PROTEINS
217
lipid
bilayer
6 nm
Figure 10–10 Schematic diagram of AE1,
represented as a cylinder, in the plasma
membrane (Problem 10–49). AE1 was
originally identiied as a prominent band
(“band 3”) after the membrane proteins
from red cell ghosts were separated by
sDs polyacrylamide-gel electrophoresis;
it is often referred to as band 3 in the
literature.
AE1
B. Calculate the fraction of the plasma membrane that is occupied by AE1.
Assume that AE1 is a cylinder 3 nm in radius and 10 nm in height and is
oriented in the membrane as shown in Figure 10–10. he total surface
area of a red cell is 108 nm2.
DATA HANDLING
10–50
Look carefully at the transmembrane anion exchanger (AE1) in
Figure 10–11A. Imagine that you could mark all the AE1 proteins speciically with a luorescent group and measure their mobility by luorescence recovery after photobleaching (FRAP). You photobleach a small
spot on the membrane and then measure the increase in luorescence in
that spot as luorescent AE1 molecules difuse into it from neighboring
regions of the membrane. Sketch the recovery curve you would expect to
see with time (Figure 10–11B).
MEDICAL LINKS
Pythagoras forbade his followers to eat fava beans. Beyond the political
implications (Greeks voted with beans), there turns out to be a rational
basis for this proscription. In the Middle Eastern population, defective
forms of the gene encoding glucose 6-phosphate dehydrogenase (G6PD)
are common. hese mutant forms of the gene typically reduce G6PD
activity to about 10% of normal. hey have been selected for in the Middle
East, and in other areas of the world where malaria is common, because
they aford protection against the malarial parasite. G6PD controls the
irst step in the pathway for NADPH production. A lower-than-normal
level of NADPH in red blood cells creates an environment unfavorable
for growth of the protist Plasmodium falciparum, which causes malaria.
Although somewhat protected against malaria, G6PD-deicient individuals occasionally have other problems. NADPH is the principal agent
required to keep the red cell cytosol in a properly reduced state, constantly converting transient disulide bonds (–S–S–) back to sulfhydryls
(–SH HS–). When a G6PD-deicient individual eats raw or undercooked
fava beans, an oxidizing substance in the beans overwhelms the reducing capacity of the red cells, leading to a severe—sometimes life-threatening—hemolytic anemia. How do you suppose eating fava beans leads
to anemia?
MCAT STYLE
Passage 1 (Questions 10–52 to 10–56)
Cholesterol is transported in the bloodstream in lipoprotein particles. here are
several classes of lipoprotein particles, but one of the most important classes is
called low-density lipoprotein (LDL) particles. LDL particles are composed of a
single molecule of a large protein called apolipoprotein, as well as thousands of
(A)
AE1
AE1
100 nm
(B)
BLEACH
fluorescence
10–51
? RECOVERY
time
Figure 10–11 Mobility of membrane
proteins (Problem 10–50). (a) Model of
red blood cell membrane. (B) recovery of
luorescence after photobleaching a red
cell plasma membrane containing ae1
protein tagged with a luorescent group.
Chapter 10: Membrane Structure
218
cholesterol molecules and phospholipids. Part of the apolipoprotein forms the
surface of the particle and faces the aqueous phase, while other parts face the
hydrophobic interior of the particle where they interact with cholesterol and the
hydrophobic tails of the phospholipids. When cells need additional cholesterol,
they produce a receptor protein on the surface of their plasma membrane that
binds to apolipoprotein and initiates uptake of LDL particles into the cell via a
process called endocytosis.
What is the most likely reason that LDL particles are used to transport
cholesterol in the bloodstream?
A. Cholesterol in LDL particles easily inserts into lipid bilayers.
B. Cholesterol is largely insoluble in aqueous solutions.
C. Cholesterol is synthesized within LDL particles.
D. LDL particles can form lipid rafts in the lipid bilayer.
10–52
What best describes the class of protein that apolipoprotein belongs to?
A. Amphiphilic proteins
B. Cell-surface receptors
C. Membrane-associated proteins
D. Multipass transmembrane proteins
10–53
What is the most likely location of phospholipids in LDL particles?
A. Bound to hydrophilic domains of apolipoprotein
B. In the lipid bilayer that surrounds LDL particles
C. On the surface, interacting with the aqueous environment
D. Within the hydrophobic core of the particle
10–54
What kind of molecule on the surface of a cell would be most likely to
serve as a receptor for LDL particles?
A. A cholesterol-binding protein
B. A glycolipid
C. A phospholipid
D. A transmembrane protein
10–55
What conditions would be likely to trigger increased uptake of LDL particles by cells?
A. Increased glycolipid synthesis
B. Increased plasma membrane growth
C. Increased protein secretion
D. Increased protein synthesis
10–56
Chapter 11
219
Membrane Transport of Small
Molecules and the Electrical
Properties of Membranes
11
PRINCIPLES OF MEMBRANE TRANSPORT
IN THIS CHAPTER
TERMS TO LEARN
active transport
channel
electrochemical gradient
membrane transport protein
passive transport
transporter
DEFINITIONS
Match each deinition below with its term from the list above.
11–1
An aqueous pore in a lipid membrane, with walls made of protein,
through which selected ions or molecules can pass.
11–2
he movement of a small molecule or ion across a membrane due to a
diference in concentration or electrical charge.
11–3
General term for a membrane-embedded protein that serves as a carrier
of ions or small molecules from one side of the membrane to the other.
11–4
Movement of a molecule across a membrane that is driven by ATP
hydrolysis or other form of metabolic energy.
11–5
Driving force for ion movement that is due to diferences in ion concentration and electrical charge on either side of the membrane.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
11–6
he plasma membrane is highly impermeable to all charged molecules.
11–7
Transport by transporters can be either active or passive, whereas transport by channels is always passive.
THOUGHT PROBLEMS
11–8
Order the molecules on the following list according to their ability to
difuse through a lipid bilayer, beginning with the one that crosses the
bilayer most readily. Explain your order.
1. Ca2+
2. CO2
3. Ethanol
4. Glucose
5. RNA
6. H2O
11–9
Why are the maximum rates of transport by transporters and channels
thought to be so diferent?
CHAPTER
PRINCIPLES OF MEMBRANE
TRANSPORT
TRANSPORTERS AND ACTIVE
MEMBRANE TRANSPORT
CHANNELS AND THE
ELECTRICAL PROPERTIES OF
MEMBRANES
220
Chapter 11: Membrane Transport of Small Molecules and the Electrical Properties of Membranes
11–10
A simple enzyme reaction can be represented by the equation
E+S
ES → E + P
where E is the enzyme, S is the substrate, P is the product, and ES is the
enzyme–substrate complex.
A. Write a corresponding equation describing a transporter (T) that mediates transport of a solute (S) down its concentration gradient.
B. he Michaelis–Menten equation for the simple enzyme reaction above is
rate = Vmax
[S]
[S] + Km
where “rate” is the initial rate of the reaction, Vmax is the maximum rate
of the enzyme-catalyzed reaction, [S] is the concentration of substrate,
and Km is the Michaelis constant. Write the corresponding Michaelis–
Menten equation for the process of solute transport by a transporter.
What do rate, Vmax, and Km mean in the equation for transport?
C. Would these equations provide an appropriate description for channels?
Why or why not?
11–11
Suppose a membrane contains a single passive transporter with a Km of
0.1 mM for its solute. How efective would the transporter be at equalizing the concentrations of solute across the membrane if the starting
concentrations were 0.01 mM inside and 0.05 mM outside? What if the
concentrations were 100 mM inside and 500 mM outside?
11–12
How is it possible for some molecules to be at equilibrium across a biological membrane and yet not be at the same concentration on both
sides?
CALCULATIONS
Brain cells, which depend on glucose for energy, use the glucose transporter GLUT3, which has a Km of 1.5 mM. Liver cells, which store glucose
(as glycogen) after a meal and release glucose between meals, use the
glucose transporter GLUT2, which has a Km of 15 mM.
A. Calculate the rate (as a percentage of Vmax) of glucose uptake in brain
cells and in liver cells at circulating glucose concentrations of 3 mM (starvation conditions), 5 mM (normal levels), and 7 mM (after a carbohydrate-rich meal). Rearranging the Michaelis–Menten equation gives
11–13
rate
[S]
=
Vmax [S] + Km
B. Although the concentration of glucose in the general circulation normally doesn’t rise much above 7 mM, the liver is exposed to much higher
concentrations after a meal. he intestine delivers glucose into the portal
circulation, which goes directly to the liver. In the portal circulation, the
concentration of glucose can be as high as 15 mM. At what fraction of the
maximum rate (Vmax) do liver cells import glucose at this concentration?
C. Do these calculations it with the physiological functions of brain and
liver cells? Why or why not?
Cells use transporters to move nearly all metabolites across membranes.
But how much faster is a transporter than simple difusion? here is suficient information available for glucose transporters to make a comparison. he normal circulating concentration of glucose in humans is 5
mM, whereas the intracellular concentration is usually very low. (For this
problem assume the internal concentration of glucose is 0 mM.)
A. At what rate (molecules/sec) would glucose difuse into a cell if
there were no transporter? he permeability coeicient for glucose is
11–14
PRINCIPLES OF MEMBRANE TRANSPORT
221
3 × 10–8 cm/sec. Assume a cell is a sphere with a diameter of 20 μm. he
rate of difusion equals the concentration diference multiplied by the
permeability coeicient and the total surface area of the cell (surface
area = 4πr2). (Remember to convert everything to compatible units so
that the rate is molecules/sec.)
B. If in the same cell there are 105 GLUT3 molecules (Km = 1.5 mM) in
the plasma membrane, each of which can transport glucose at a maximum rate of 104 molecules per second, at what rate (molecules/sec) will
glucose enter the cell? How much faster is transporter-mediated uptake
of glucose than entry by simple difusion?
DATA HANDLING
11–15
Cytochalasin B, which is often used as an inhibitor of actin-based motility
systems, is also a very potent competitive inhibitor of d-glucose uptake
into mammalian cells. When red blood cell ghosts (red cells, emptied of
their cytoplasm) are incubated with 3H-cytochalasin B and then irradiated with ultraviolet light, the cytochalasin becomes cross-linked to
the glucose transporter, GLUT1. Cytochalasin is not cross-linked to the
transporter if an excess of d-glucose is present during the labeling reaction; however, an excess of L-glucose (which is not transported) does
not interfere with labeling. Why does an excess of d-glucose, but not
L-glucose, prevent cross-linking of cytochalasin to GLUT1?
Insulin is a small protein hormone that binds to receptors in the plasma
membranes of many cells. In fat cells, this binding dramatically increases
the rate of uptake of glucose into the cells. he increase occurs within
minutes and is not blocked by inhibitors of protein synthesis or of glycosylation. hese results suggest that insulin increases the activity of the
glucose transporter, GLUT4, in the plasma membrane, without increasing the total number of GLUT4 molecules in the cell.
he two experiments described below suggest a possible mechanism
for this insulin efect. In the irst experiment, the initial rate of glucose
uptake in control and insulin-treated cells was measured, with the results
shown in Figure 11–1. In the second experiment, the concentration of
GLUT4 in fractionated membranes from control and insulin-treated cells
was measured, using the binding of radioactive cytochalasin B as the
assay (see Problem 11–15), as shown in Table 11–1.
A. Deduce the mechanism by which glucose transport through GLUT4
increases in insulin-treated cells.
B. Does insulin stimulation alter either the Km or the Vmax of GLUT4? How
can you tell from these data?
11–16
TABLE 11–1 Amount of GLUT4 associated with the plasma membrane
and internal membranes in the presence and absence of insulin
(Problem 11–16).
Bound 3H-cytochalasin B (cpm/mg protein)
Membrane fraction
Untreated cells
(– insulin)
Treated cells
(+ insulin)
Plasma membrane
890
4480
Internal membranes
4070
80
rate of glucose uptake
(arbitrary units)
30
+ insulin
20
10
– insulin
0
0
10
glucose (mM)
20
Figure 11–1 Rate of glucose uptake into
cells in the presence and absence of
insulin (Problem 11–16).
222
Chapter 11: Membrane Transport of Small Molecules and the Electrical Properties of Membranes
TRANSPORTERS AND ACTIVE MEMBRANE
TRANSPORT
TERMS TO LEARN
ABC transporter
antiporter
Ca2+-pump (Ca2+ ATPase)
multidrug resistance (MDR) protein
Na+-K+ pump (Na+-K+ ATPase)
P-type pump
symporter
transcellular transport
uniporter
V-type pump
DEFINITIONS
Match each deinition below with its term from the list above.
11–17
Large superfamily of membrane transport proteins that use the energy
of ATP hydrolysis to transfer peptides and a variety of small molecules
across membranes.
11–18
Type of ABC transporter protein that can pump hydrophobic drugs (such
as some anticancer drugs) out of the cytoplasm of eukaryotic cells.
11–19
Membrane carrier protein that transports two diferent ions or small
molecules across a membrane in opposite directions, either simultaneously or in sequence.
11–20
Transport of solutes across an epithelium, by means of membrane transport proteins in the apical and basal surfaces of the epithelial cells.
11–21
Carrier protein that transports two types of solute across the membrane
in the same direction.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
11–22
A symporter would function as an antiporter if its orientation in the
membrane were reversed; that is, if the portion of the protein normally
exposed to the cytosol faced the outside of the cell instead.
11–23
he co-transport of Na+ and a solute into a cell, which harnesses the
energy in the Na+ gradient, is an example of primary active transport.
11–24
In response to depolarization of the muscle cell plasma membrane, the
Ca2+-pumps in the sarcoplasmic reticulum (SR) use the energy of ATP
hydrolysis to move Ca2+ from the lumen of the SR to the cytosol to initiate
muscle contraction.
THOUGHT PROBLEMS
11–25
Which of the ions listed in Table 11–2 could be used to drive an electrically neutral coupled transport of a solute across the plasma membrane?
Indicate the direction of movement of the listed ions (inward or outward)
and indicate what sort of ion would be co-transported to preserve electrical neutrality.
Incidentally, there is a glaring intracellular deiciency of anions relative to cations in Table 11–2, yet cells are electrically neutral. What anions
do you suppose are missing from this table?
11–26
A transmembrane protein has the following properties: it has two binding sites, one for solute A and one for solute B. he protein can undergo a
conformational change to switch between two states: either both binding
sites are exposed exclusively on one side of the membrane, or both are
TRANSPORTERS AND ACTIVE MEMBRANE TRANSPORT
223
TABLE 11–2 A comparison of ion concentrations inside and outside a typical
mammalian cell (Problem 11–25).
Component
Intracellular concentration
(mM)
Extracellular concentration
(mM)
Na+
5–15
145
K+
140
5
Mg2+
0.5
1–2
Ca2+
10–4
1–2
H+
7 × 10–5 (10–7.2 M or pH 7.2)
4 × 10–5 (10–7.4 M or pH 7.4)
5–15
110
Cations
Anions
Cl–
exposed exclusively on the other side of the membrane. he protein can
switch between the two conformational states only if both binding sites
are occupied or if both binding sites are empty, but cannot switch if only
one binding site is occupied.
A. What kind of a transporter do these properties deine?
B. Do you need to specify any additional properties to turn this protein into
a transporter that couples the movement of solute A up its concentration
gradient to the movement of solute B down its electrochemical gradient?
C. Write a set of rules like those in the body of this problem that deines the
properties of an antiporter.
11–27
A model for a uniporter that could mediate passive transport of glucose
down its concentration gradient is shown in Figure 11–2. How would you
need to change the diagram to convert the transporter into a pump that
transports glucose up its concentration gradient by hydrolyzing ATP?
Explain the need for each of the steps in your new illustration.
glucose
gradient
Ion transporters are “linked” together—not physically, but as a consequence of their actions. For example, cells can raise their intracellular
pH, when it becomes too acidic, by exchanging external Na+ for internal H+, using a Na+–H+ antiporter. he change in internal Na+ is then
redressed using the Na+-K+ pump.
A. Can these two transporters, operating together, normalize both the H+
and the Na+ concentrations inside the cell?
B. Does the linked action of these two pumps cause imbalances in either
the K+ concentration or the membrane potential?
11–28
Figure 11–2 Hypothetical model
showing how a conformational change
in a transporter could mediate passive
transport of glucose (Problem 11–27). The
transition between the two conformational
states is proposed to occur randomly and
to be completely reversible, regardless of
binding-site occupancy.
Chapter 11: Membrane Transport of Small Molecules and the Electrical Properties of Membranes
224
Figure 11–3 The Na+-k+ pump (Problem
11–29).
Na+
ADP
ATP
P
P
Na+
K+
P
K+
P
You have prepared lipid vesicles (spherical lipid bilayers) that contain
Na+-K+ pumps as the sole membrane protein. Assume for the sake of
simplicity that each pump transports one Na+ one way and one K+ the
other way in each pumping cycle, as illustrated in Figure 11–3. All of
the Na+-K+ pumps are oriented so that the portion of the molecule that
normally faces the cytosol faces the outside of the vesicle. Predict what
would happen under each of the following conditions.
A. he solution inside and outside the vesicles contains both Na+ and K+
ions, but no ATP.
B. he solution inside the vesicles contains both Na+ and K+ ions; the solution outside also contains both ions, as well as ATP.
C. he solution inside contains Na+; the solution outside contains Na+ and
ATP.
D. he solution is as in B, but the Na+-K+ pump molecules are randomly
oriented, some facing one direction, some the other.
11–29
CALCULATIONS
11–30
Microvilli increase the surface area of intestinal cells, providing more
eicient absorption of nutrients. Microvilli are shown in proile and cross
section in Figure 11–4. From the dimensions given in the igure, estimate
the increase in surface area that microvilli provide (for the portion of the
plasma membrane in contact with the lumen of the gut) relative to the
corresponding surface of a cell with a “lat” plasma membrane.
1 µm
0.1 µm
Figure 11–4 Microvilli of intestinal
epithelial cells in proile and cross
section (Problem 11–30).
TRANSPORTERS AND ACTIVE MEMBRANE TRANSPORT
11–31
225
How much energy does it take to pump substances across membranes?
Or, to put it another way, since active transport is usually driven directly
or indirectly by ATP, how steep a gradient can ATP hydrolysis maintain
for a particular solute? For transport into the cell, the free-energy change
(∆Gin) per mole of solute moved across the plasma membrane is
∆ Gin = –2.3RT log
Co
+ zFV
Ci
where R = the gas constant, 8.3 × 10–3 kJ/K mole
T = the absolute temperature in K (37°C = 310 K)
Co = solute concentration outside the cell
Ci = solute concentration inside the cell
z = the valence (charge) on the solute
F = Faraday’s constant, 96 kJ/V mole
V = the membrane potential in volts (V)
Since ∆Gin = –∆Gout, the free-energy change for transport out of the cell is
∆ Gout = 2.3RT log
Co
– zFV
Ci
At equilibrium, where ∆G = 0, the equations can be rearranged to the
more familiar form known as the Nernst equation.
V = 2.3
RT
zF
log
Co
Ci
For the questions below, assume that hydrolysis of ATP to ADP and
Pi proceeds with a ∆G of –50 kJ/mole; that is, ATP hydrolysis can drive
active transport with a ∆G of +50 kJ/mole. Assume that V is –60 mV.
A. What is the maximum concentration gradient that can be achieved by
the ATP-driven active transport into the cell of an uncharged molecule
such as glucose, assuming that 1 ATP is hydrolyzed for each solute molecule that is transported?
B. What is the maximum concentration gradient that can be achieved by
active transport of Ca2+ from the inside to the outside of the cell? How
does this maximum compare with the actual concentration gradient
observed in mammalian cells (see Table 11–2)?
C. Calculate how much energy it takes to drive the Na+-K+ pump. his
remarkable molecular device transports ive ions for every molecule of
ATP that is hydrolyzed: 3 Na+ out of the cell and 2 K+ into the cell. he
pump typically maintains internal Na+ at 10 mM, external Na+ at 145 mM,
internal K+ at 140 mM, and external K+ at 5 mM. As shown in Figure 11–5,
Na+ is transported against the membrane potential, whereas K+ is transported with it. (he ∆G for the overall reaction is equal to the sum of the
∆G values for transport of the individual ions.)
D. How eicient is the Na+-K+ pump? hat is, what fraction of the energy
available from ATP hydrolysis is used to drive transport?
DATA HANDLING
11–32
If you have ever used the standard probe on a pH meter, you may well
wonder how pH could possibly be measured in the tiny volumes inside
cellular compartments. he recent development of pH-sensitive luorophores has simpliied this diicult task immensely. One such luorescent
indicator is a hydrophobic ester of SNARF-1, which can enter cells by
passive difusion and then is trapped inside after intracellular enzymes
hydrolyze the ester bonds to liberate SNARF-1 (Figure 11–6). SNARF-1
absorbs light at 488 nm and emits luorescent light with peaks at 580
nm and 640 nm. Emission spectra for SNARF-1 at pH 6.0 and pH 9.0 are
OUTSIDE
CELL
INSIDE
CELL
+
K+
+
+
+
+
+
K+
Na+
+
+
+
+
+
+
+
+
Na+
+
plasma
membrane
Figure 11–5 Na+ and k+ gradients and
direction of pumping across the plasma
membrane (Problem 11–31). Large letters
symbolize high concentrations and small
letters symbolize low concentrations.
Both Na+ and k+ are pumped against
chemical concentration gradients;
however, Na+ is pumped up the electrical
gradient, whereas k+ is pumped down
the electrical gradient.
Chapter 11: Membrane Transport of Small Molecules and the Electrical Properties of Membranes
226
O
(CH3)2N
O
O
O
C
O
C
O–
O
–
O
SNARF-1 ester
O
(CH3)2N
O
C
O
C
O
(CH3)2N
O
pK = 7.5
HYDROLYSIS
C
RO
O
(CH3)2N
O
O–
OH
OR
–
O
C
O
SNARF-1 (HA)
–
O
C
–
O SNARF-1 (A )
C
O
SNARF-1 (A–)
dominant resonance structure
Figure 11–6 Structure of the ester and free forms of SNARF-1 (Problem 11–32). The blocking ester groups are shown as
R. The acid (HA) and salt (A–) forms of SNARF-1 are indicated. The two different resonance structures for the salt form of
SNARF-1 are shown in brackets.
shown in Figure 11–7. he pK of SNARF-1 is 7.5.
A. Explain why the ester of SNARF-1 difuses through membranes, whereas
the cleaved form stays inside cells.
B. Why do you think there are two peaks of luorescence (at 580 nm and at
640 nm) that change so dramatically in intensity with a change in pH (see
Figure 11–7)? What features of SNARF-1 might be important in this?
C. What forms of SNARF-1 are present at pH 6.0 and what are their relative
proportions? At pH 9.0? he Henderson–Hasselbalch equation describing the dissociation of a weak acid is pH = pK + log ([salt]/[acid]).
D. Sketch an approximate curve for the SNARF-1 emission spectrum inside
a cell at pH 7.2. (All such curves pass through the point where the two
curves in Figure 11–7 cross.)
E. Why do you suppose indicators such as SNARF-1 that have emission
spectra with two peaks are preferred to those that have a single peak?
MEDICAL LINKS
CO2 is removed from the body through the lungs in a process that is
mediated by red blood cells, as summarized in Figure 11–8. Transport of
CO2 is coupled to the transport of O2 through hemoglobin. Upon release
of O2 in the tissues, hemoglobin undergoes a conformational change that
raises the pK of a histidine side chain, allowing it to bind an H+, which is
generated during hydration of CO2 by the action of the enzyme carbonic
anhydrase. his process occurs in reverse in the lungs when O2 is bound
to hemoglobin.
A. To what extent does the intracellular pH of the red blood cell vary during
its movement from the tissues to the lungs and back, and why is this so?
B. In what form, and where, is the CO2 during its movement from the tissues
to the lungs?
C. How is it that the Cl––HCO3– exchanger operates in one direction in the
tissues and in the opposite direction in the lungs?
11–33
11–34
A rise in the intracellular Ca2+ concentration causes muscle cells to contract. In addition to an ATP-driven Ca2+-pump, heart muscle cells, which
contract quickly and regularly, have an antiporter that exchanges Ca2+ for
extracellular Na+ across the plasma membrane. his antiporter rapidly
pumps most of the entering Ca2+ ions back out of the cell, allowing the
cell to relax. Ouabain and digitalis, drugs that are used in the treatment of
patients with heart disease, make the heart contract more strongly. Both
drugs function by partially inhibiting the Na+-K+ pump in the membrane
of the heart muscle cell. Can you propose an explanation for the efects
of these drugs in patients? What will happen if too much of either drug is
taken?
pH 9.0
fluorescence
pH 6.0
550
600
650
700
750
wavelength (nm)
Figure 11–7 Emission spectra of
SNARF-1 at pH 6.0 and pH 9.0 (Problem
11–32). SNARF-1 in solution at pH 6.0
or pH 9.0 was excited by light at 488 nm
and the intensity of luorescence was
determined at wavelengths from 550 nm
to 750 nm.
CHANNELS AND THE ELECTRICAL PROPERTIES OF MEMBRANES
IN TISSUES
low O2, high CO2
CO2
CO2 + H2O
HCO3– + H+
O2
227
IN LUNGS
high O2, low CO2
CO2
N
CO2 + H2O
HCO3– + H+
Cl–
HCO3–
CHANNELS AND THE ELECTRICAL PROPERTIES
OF MEMBRANES
TERMS TO LEARN
acetylcholine receptor
action potential
adaptation
AMPA receptor
aquaporin (water channel)
axon
Ca2+-activated K+ channel
channelrhodopsin
delayed K+ channel
dendrite
depolarization
excitatory neurotransmitter
glial cell
inhibitory neurotransmitter
initial segment
ion channel
K+ leak channel
long-term depression (LDP)
long-term potentiation (LTP)
mechanosensitive channel
membrane potential
H+
N
N
N
HCO3–
O2
metabotropic receptor
myelin sheath
Nernst equation
neuromuscular junction
neuron (nerve cell)
neurotransmitter
NMDA receptor
oligodendrocyte
optogenetics
patch-clamp recording
resting membrane potential
rapidly inactivating K+ channel
Schwann cell
selectivity filter
synapse
synaptic plasticity
transmitter-gated ion channel
voltage-gated cation channel
voltage-gated K+ channel
voltage-gated Na+ channel
DEFINITIONS
Match each deinition below with its term from the list above.
11–35
he adjustment in sensitivity of a cell or organism following repeated
stimulation that allows a response even when there is a high background
level of stimulation.
11–36
he long-lasting increase (days to weeks) in the sensitivity of certain synapses in the hippocampus that is induced by a short burst of repetitive
iring in the presynaptic neurons.
11–37
Rapid, transient, self-propagating electrical signal in the plasma membrane of a cell such as a neuron or muscle cell: a nerve impulse.
11–38
Quantitative expression that relates the equilibrium ratio of concentrations of an ion on either side of a permeable membrane to the voltage
diference across the membrane.
11–39
A photosensitive ion channel that opens in response to light.
11–40
Voltage diference across a membrane due to the slight excess of positive ions on one side and of negative ions on the other (typically –60 mV,
inside negative, for an animal cell).
Cl–
Figure 11–8 Red blood cell-mediated
transport of Co2 from the tissues to the
lungs (Problem 11–33). low and high o2
and Co2 refer to their concentrations, or
partial pressures.
228
Chapter 11: Membrane Transport of Small Molecules and the Electrical Properties of Membranes
11–41
General term for a membrane protein that selectively allows cations
such as Na+ to cross a membrane in response to changes in membrane
potential.
11–42
hat part of an ion channel structure that determines which ions it can
transport.
11–43
Insulating layer of specialized cell membrane wrapped around vertebrate axons.
11–44
A K+-transporting ion channel in the plasma membrane of animal cells
that remains open even in a “resting” cell.
11–45
Long, thin nerve cell process capable of rapidly conducting nerve
impulses over long distances so as to deliver signals to other cells.
11–46
Transmembrane protein complex that forms a water-illed channel
across the lipid bilayer through which speciic inorganic ions can difuse
down their electrochemical gradients.
11–47
Specialized junction between a nerve cell and another cell, across which
the nerve impulse is transferred, usually by a neurotransmitter, which is
secreted by the nerve cell and difuses to the target cell.
11–48
Small signaling molecule such as acetylcholine, glutamate, GABA, or glycine, secreted by a nerve cell at a chemical synapse to signal to the postsynaptic cell.
11–49
Technique in which the tip of a small glass electrode is sealed onto an
area of cell membrane, thereby making it possible to record the low of
current through individual ion channels.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
11–50
Transporters saturate at high concentrations of the transported molecule
when all their binding sites are occupied; channels, on the other hand,
do not bind the ions they transport and thus the lux of ions through a
channel does not saturate.
11–51
he membrane potential arises from movements of charge that leave ion
concentrations practically unafected, causing only a very slight discrepancy in the number of positive and negative ions on the two sides of the
membrane.
11–52
he aggregate current crossing the membrane of an entire cell indicates
the degree to which individual channels are open.
11–53
Transmitter-gated ion channels open in response to speciic neurotransmitters in their environment but are insensitive to the membrane potential; therefore, they cannot by themselves (in the absence of ligand) generate an action potential.
11–54
When an action potential depolarizes the muscle cell membrane, the
Ca2+-pump is responsible for pumping Ca2+ from the sarcoplasmic reticulum into the cytosol to initiate muscle contraction.
THOUGHT PROBLEMS
11–55
According to Newton’s laws of motion, an ion exposed to an electric ield
in a vacuum would experience a constant acceleration from the electric
driving force, just as a falling body in a vacuum constantly accelerates
due to gravity. In water, however, an ion moves at constant velocity in an
electric ield. Why do you suppose that is?
CHANNELS AND THE ELECTRICAL PROPERTIES OF MEMBRANES
11–56
What two properties distinguish an ion channel from a simple aqueous
pore?
229
H+
You have prepared lipid vesicles that contain molecules of the K+ leak
channel, all oriented so that their cytosolic surface faces the outside of
the vesicle. Predict how K+ ions will move under the following conditions
and what sort of membrane potential will develop.
A. Equal concentrations of K+ ion are present inside and outside the vesicle.
B. K+ ions are present only inside the vesicle.
C. K+ ions are present only outside the vesicle.
11–57
11–58
If a frog egg and a red blood cell are placed in pure water, the red blood
cell will swell and burst, but the frog egg will remain intact. Although a
frog egg is about one million times larger than a red cell, they both have
nearly identical internal concentrations of ions so that the same osmotic
forces are at work in each. Why do you suppose red blood cells burst in
water, while frog eggs do not?
11–59
Aquaporins allow water to move across a membrane, but prevent the
passage of ions. How does the structure of the pore through which the
water molecules move prevent passage of ions such as K+, Na+, Ca+, and
Cl–? H+ ions present a diferent problem because they move by relay
along a chain of hydrogen-bonded water molecules (Figure 11–9). How
does the pore prevent the relay of H+ ions across the membrane?
11–60
Explain in 100 words or fewer how an action potential is passed along an
axon.
11–61
he neurotransmitter acetylcholine is made in the cytosol and then
transported into synaptic vesicles, where its concentration is more than
100-fold higher than in the cytosol. Synaptic vesicles isolated from neurons can take up additional acetylcholine if it is added to the solution in
which they are suspended, but only in the presence of ATP. Na+ ions are
not required for acetylcholine uptake, but, curiously, raising the pH of
the solution in which the synaptic vesicles are suspended increases acetylcholine uptake. Furthermore, transport is inhibited in the presence of
drugs that make the membrane permeable to H+ ions. Suggest a mechanism that is consistent with all these observations.
11–62
Excitatory neurotransmitters open Na+ channels, while inhibitory neurotransmitters open either Cl– or K+ channels. Rationalize this observation
in terms of the efects of these ions on the iring of an action potential.
11–63
Acetylcholine-gated cation channels do not discriminate among Na+, K+,
and Ca2+ ions, allowing all to pass through freely. How is it, then, that
when acetylcholine receptors in muscle cells open there is a large net
inlux principally of Na+?
CALCULATIONS
11–64
In a subset of voltage-gated K+ channels, the N-terminus of each subunit
acts like a tethered ball that occludes the cytoplasmic end of the pore
soon after it opens, thereby inactivating the channel. his “ball-andchain” model for the rapid inactivation of voltage-gated K+ channels
has been elegantly supported for the shaker K+ channel from Drosophila melanogaster. (he shaker K+ channel in Drosophila is named after
a mutant form that causes excitable behavior—even anesthetized lies
keep twitching.) Deletion of the N-terminal amino acids from the normal shaker channel gives rise to a channel that opens in response to
membrane depolarization but stays open (Figure 11–10A; 0 μM) instead
of rapidly closing as the normal channel does. A peptide (MAAVAGLYGLGEDRQHRKKQ) that corresponds to the deleted N-terminus can
partially inactivate the open channel at 50 μM and completely inactivate
it at 100 μM (Figure 11–10A).
H+
Figure 11–9 Rapid diffusion of H+ ions
by a molecular relay system involving
the making and breaking of hydrogen
bonds between adjacent water molecules
(Problem 11–59).
Chapter 11: Membrane Transport of Small Molecules and the Electrical Properties of Membranes
230
(A)
(B)
0 µM
21.4 nm
50 µM
100 pA
10 msec
100 µM
Is the concentration of free peptide (100 μM) required to inactivate
the defective K+ channel anywhere near the local concentration of the
tethered ball on a normal channel? Assume that the tethered ball can
explore a hemisphere [volume = (2/3)πr3] with a radius of 21.4 nm, which
is the length of the polypeptide “chain” (Figure 11–10B). Calculate the
concentration for one ball in this hemisphere. How does that value compare with the concentration of free peptide needed to inactivate the
channel?
11–65
If the resting membrane potential of a cell is –70 mV and the thickness of
the lipid bilayer is 4.5 nm, what is the strength of the electric ield across
the membrane in V/cm? What do you suppose would happen if you
applied this voltage to two metal electrodes separated by a 1 cm air gap?
11–66
he squid giant axon occupies a unique position in the history of our
understanding of cell membrane potentials and nerve action. Its large
size (0.2–1.0 mm in diameter and 5–10 cm in length) allowed electrodes,
large by modern standards, to be inserted so that intracellular voltages
could be measured. When an electrode is stuck into an intact giant axon,
the membrane potential registers –70 mV. When the axon, suspended in
a bath of seawater, is stimulated to conduct a nerve impulse, the membrane potential changes transiently from –70 mV to +40 mV.
he Nernst equation relates equilibrium ionic concentrations to the
membrane potential.
V = 2.3
RT
C
log o
zF
Ci
For univalent ions and at 20°C (293 K),
V = 58 mV × log
Co
Ci
A. Using this equation, calculate the potential across the resting membrane
(1) assuming that it is due solely to K+ and (2) assuming that it is due
solely to Na+. (he Na+ and K+ concentrations in the axon cytosol and
in seawater are given in Table 11–3.) Which calculation is closer to the
measured resting potential? Which calculation is closer to the measured
action potential? Explain why these assumptions approximate the measured resting and action potentials.
B. If the solution bathing the squid giant axon is changed from seawater to
artiicial seawater in which NaCl is replaced with choline chloride, there
is no efect on the resting potential, but the nerve no longer generates an
action potential upon stimulation. What would you predict would happen to the magnitude of the action potential if the concentration of Na+
TABLE 11–3 Ionic composition of seawater and of the cytosol in the
squid giant axon (Problem 11–66).
Ion
Cytosol
Seawater
Na+
65 mM
430 mM
K+
344 mM
9 mM
Figure 11–10 Inactivation of voltagegated k+ channels (Problem 11–64).
(A) Patch-clamp recording of a defective
shaker k+ channel in the absence and
presence of inactivating peptide. Current
through the channel is indicated in
picoamps (pA). (B) A “ball” tethered by a
“chain” to a normal channel.
CHANNELS AND THE ELECTRICAL PROPERTIES OF MEMBRANES
231
in the external medium were reduced to a half or a quarter of its normal
value, using choline chloride to maintain osmotic balance?
he number of Na+ ions entering the squid giant axon during an action
potential can be calculated from theory. Because the cell membrane
separates positive and negative charges, it behaves like a capacitor. From
the known capacitance of biological membranes, the number of ions that
enter during an action potential can be calculated. Starting from a resting
potential of –70 mV, it can be shown that 1.1 × 10–12 moles of Na+ must
enter the cell per cm2 of membrane during an action potential.
To determine experimentally the number of entering Na+ ions during
an action potential, a squid giant axon (1 mm in diameter and 5 cm in
length) was suspended in a solution containing radioactive Na+ (speciic
activity = 2 × 1014 cpm/mole) and a single action potential was propagated down its length. When the cytoplasm was analyzed for radioactivity, a total of 340 cpm were found to have entered the axon.
A. How well does the experimental measurement match the theoretical calculation?
B. How many moles of K+ must cross the membrane of the axon, and in
which direction, to reestablish the resting potential after the action
potential is over?
C. Given that the concentration of Na+ inside the axon is 65 mM, calculate
the fractional increase in internal Na+ concentration that results from the
passage of a single action potential down the axon.
D. At the other end of the spectrum of nerve sizes are small dendrites about
0.1 μm in diameter. Assuming the same length (5 cm), the same internal
Na+ concentration (65 mM), and the same resting and action potentials
as for the squid giant axon, calculate the fractional increase in internal
Na+ concentration that would result from the passage of a single action
potential down a dendrite.
E. Is the Na+-K+ pump more important for the continuing performance of a
giant axon, or of a dendrite?
11–67
Acetylcholine-gated cation channels at the neuromuscular junction
open in response to acetylcholine released by the nerve terminal and
allow Na+ ions to enter the muscle cell, which causes membrane depolarization and ultimately leads to muscle contraction.
A. Patch-clamp measurements show that young rat muscles have cation
channels that respond to acetylcholine (Figure 11–11). How many kinds
of channel are there? How can you tell?
B. For each kind of channel, calculate the number of ions that enter in one
millisecond. (One ampere is a current of one coulomb per second; one
pA equals 10–12 ampere. An ion with a single charge such as Na+ carries a
charge of 1.6 × 10–19 coulomb.)
11–68
Figure 11–11 Patch-clamp measurements
of acetylcholine-gated cation channels in
young rat muscle (Problem 11–68).
2 pA
40 msec
DATA HANDLING
11–69
he shaker K+ channel in Drosophila opens in response to membrane
depolarization and then rapidly inactivates via a ball-and-chain mechanism (see Problem 11–64). he shaker K+ channel assembles as a tetramer
composed of four subunits, each with its own ball and chain. Do multiple
balls in the tetrameric channel act together to inactivate the channel, or
is one ball suicient?
Chapter 11: Membrane Transport of Small Molecules and the Electrical Properties of Membranes
232
(A) MIXTURE OF K+ CHANNELS
(B) PATCH-CLAMP RECORDINGS
toxin-sensitive
current (µA)
1.0
– toxin
+ toxin
0.5
0
0
50
time (msec)
100 0
50
time (msec)
100
his question has been answered by mixing subunits from two different forms of the K+ channel: the normal subunits with balls, and scorpion-toxin-resistant subunits without balls (Figure 11–12A). Scorpion
toxin prevents the opening of normal (toxin-sensitive) K+ channels, but
not channels composed entirely of toxin-resistant subunits. Moreover,
hybrid channels containing even a single toxin-sensitive subunit fail to
open in the presence of toxin.
Mixed K+ channels were created by injecting a mixture of mRNAs for
the two types of subunit into frog oocytes. After expression, patches of
membrane containing several hundred channels were studied by patchclamp recording after subjecting the membranes to depolarization in the
absence or presence of scorpion toxin (Figure 11–12B).
A. Sketch the expected patch-clamp recordings, in the presence and
absence of toxin, for a pure population of K+ channels composed entirely
of toxin-resistant subunits without balls.
B. Sketch the expected patch-clamp recordings, in the presence and
absence of toxin, for a pure population of K+ channels composed entirely
of normal (toxin-sensitive) subunits with balls.
C. Sketch the expected patch-clamp recordings, in the presence and
absence of toxin, for a mixed population of K+ channels, 50% composed
entirely of normal (toxin-sensitive) subunits with balls and 50% composed entirely of toxin-resistant subunits without balls.
D. he key observation in Figure 11–12B is that the inal plateau values of
the currents in the absence and presence of toxin are the same. Does this
observation argue that a single ball can close a channel or that multiple
balls must act in concert? (hink about what the curves would look like if
one, two, three, or four balls were required to close a channel.)
11–70
A recording from a patch of membrane (Figure 11–13A) measured current through acetylcholine-gated ion channels, which open when acetylcholine is bound (Figure 11–13B). To obtain the recording, acetylcholine
was added to the solution in the micropipette (Figure 11–13A). Describe
what you can learn about the channels from this recording. How would
the recording difer if acetylcholine were (a) omitted or (b) added only to
the solution outside the micropipette?
11–71
One important parameter for understanding any particular membrane
transport process is to know the number of copies of the speciic transport protein present in the cell membrane. To measure the number of
(A)
Figure 11–12 Analysis of the inactivation
of the shaker k+ channel (Problem 11–69).
(A) Mixture of normal subunits with balls
(white) and toxin-resistant subunits
without balls (red). If the two subunits
were expressed in equal amounts, the
different forms of the k+ channel would
be present in the ratio 1:4:6:4:1; however,
higher levels of toxin-resistant subunit
mRNA were used, signiicantly skewing
the ratios in favor of channels with toxinresistant subunits. (B) Patch-clamp
recordings in the presence and absence
of scorpion toxin.
(B)
micropipette
ion
channels
2 pA
10 msec
CYTOSOL
Figure 11–13 Analysis of acetylcholinegated ion channels (Problem 11–70). (A) A
micropipette with a patch of membrane.
(B) Patch-clamp recording of current
through acetylcholine-gated ion channels.
CHANNELS AND THE ELECTRICAL PROPERTIES OF MEMBRANES
MEDICAL LINKS
11–72
To make antibodies against the acetylcholine receptor from the electric
organ of electric eels, you inject the puriied receptor into mice. You note
an interesting correlation: mice with high levels of antibodies against
the receptor appear weak and sluggish; those with low levels are lively.
You suspect that the antibodies against the eel acetylcholine receptors
are reacting with the mouse acetylcholine receptors, causing many of
the receptors to be destroyed. Since a reduction in the number of acetylcholine receptors is also the basis for the human autoimmune disease myasthenia gravis, you wonder whether an injection of the drug
neostigmine into the mice might give them a temporary restoration of
strength, as it does for myasthenic patients. Neostigmine inhibits acetylcholinesterase, the enzyme responsible for hydrolysis of acetylcholine in
the synaptic cleft. Sure enough, when you inject your mice with neostigmine, they immediately become very active. Propose an explanation for
how neostigmine restores temporary function to a neuromuscular synapse with a reduced number of acetylcholine receptors.
11–73
he ion channels for neurotransmitters such as acetylcholine, serotonin,
GABA, and glycine have similar overall structures. Each class comprises
an extremely diverse set of channel subtypes, with diferent ligand ainities, diferent channel conductances, and diferent rates of opening and
closing. Why is such extreme diversity a good thing from the standpoint
of the pharmaceutical industry?
MCAT STYLE
Passage 1 (Questions 11–74 to 11–76)
Numerous interesting drugs and toxins act on channels and transporters. One
such drug is scopolamine, which is used to treat vertigo, motion sickness, and
toxin bound (pmol/g wet weight)
voltage-gated Na+ channels in the rabbit vagus nerve, you use a potent
toxin, saxitoxin, a shellish poison that speciically inactivates the voltage-gated Na+ channels in these nerve cells. Assuming that each channel
binds one toxin molecule, the number of Na+ channels in a segment of
vagus nerve will be equal to the maximum number of bound toxin molecules.
You incubate identical segments of nerve for 8 hours with increasing amounts of 125I-labeled toxin. You then wash the segments to remove
unbound toxin and measure the radioactivity associated with the nerve
segments to determine the toxin-binding curve (Figure 11–14, upper
curve). You are puzzled because you expected to see binding reach a
maximum (saturate) at high concentrations of toxin; however, no distinct
end point was reached, even at higher concentrations of toxin than those
shown in the igure. After careful thought, you design a control experiment in which the binding of labeled toxin is measured in the presence
of a large molar excess of unlabeled toxin. he results of this experiment,
which are shown in the lower curve in Figure 11–14, make everything
clear and allow you to calculate the number of Na+ channels in the membrane of the vagus nerve axon.
A. Why does binding of the labeled toxin not saturate? What is the point of
the control experiment, and how does it work?
B. Given that 1 gram of vagus nerve has an axonal membrane area of 6000
cm2 and assuming that the Na+ channel is a cylinder with a diameter of
6 nm, calculate the number of Na+ channels per square micrometer of
axonal membrane and the fraction of the cell surface occupied by the
channel. (Use 100 pmol as the amount of toxin speciically bound to the
receptor.)
233
200
– unlabeled toxin
150
100
+ unlabeled toxin
50
0
0
20
40
60
80
125
I-toxin (nM)
Figure 11–14 Toxin-binding curves in
the presence and absence of unlabeled
saxitoxin (Problem 11–71).
Chapter 11: Membrane Transport of Small Molecules and the Electrical Properties of Membranes
234
smooth muscle spasms. Scopolamine was irst isolated from Jimson Weed, which
produces a beautiful white lower—often painted by Georgia O’Keefe. Jimson
Weed is also called Sacred Datura, and was used by Native Americans for religious purposes because it can produce hallucinations and time distortions. Imagine that you are the irst to purify scopolamine and are trying to determine how
it works. You ind that if you incubate isolated muscle cells with scopolamine,
subsequent addition of acetylcholine no longer causes membrane depolarization
and cell contraction, as it does in the absence of scopolamine.
Which one of the following statements best explains how scopolamine
might exert its efects?
A. It binds to the acetylcholine-gated Na+ channel and inhibits its opening.
B. It inhibits opening of the Ca2+ channel in the sarcoplasmic reticulum.
C. It inhibits transporters that import Na+ during an action potential.
D. It inhibits voltage-gated K+ channels during an action potential.
11–74
Scorpion α-toxin, a component of scorpion venom, dramatically prolongs the change in membrane potential during the iring of a nerve
impulse (Figure 11–15). Which one of the following hypotheses best
explains how α-toxin exerts its efects?
A. It accelerates the opening of voltage-gated K+ channels.
B. It opens the voltage-gated Na+ channel at a lower voltage.
C. It promotes the opening of a ligand-gated Na+ channel.
D. It slows the inactivation of the voltage-gated Na+ channel.
11–75
(A) CONTROL
(B) + TOXIN
+45
mV
0
–75
1 msec
seconds
Extracts from the African plant Stropthantus gratus were once used
to make poison arrowheads for hunting. he active compound in the
extract is extremely toxic; when concentrated and placed on an arrowhead, it can kill a hippopotamus. Imagine that you treat cells with the
compound and observe that they begin to swell and eventually burst.
Which one of the following actions would best explain the efects of the
toxic compound?
A. Closing of K+ leak channels.
B. Closing of Na+ channels.
C. Inhibition of glucose transporters.
D. Inhibition of the Na+-K+ pump.
11–76
Passage 2 (Questions 11–77 to 11–78)
Severe diarrhea associated with diseases such as cholera and dysentery was
the leading cause of infant mortality world-wide prior to 1980. Severe diarrhea
produces such a rapid luid loss that the individual becomes so dehydrated that
organs fail. When it invades the gut, Vibrio cholerae, the bacterium responsible for
cholera, produces a toxin that hyperactivates the cystic ibrosis transmembrane
regulator (CFTR). Hyperactivation of CFTR increases the movement of Cl– ions
into the lumen of the intestine, which creates an ionic imbalance. As a result,
water rushes into the intestine from surrounding tissues, leading to diarrhea and
Figure 11–15 Effects of scorpion α-toxin
on the duration of an action potential
(Problem 11–75).
MCAT STYLE
rapid water loss. Mutations that inactivate CFTR cause cystic ibrosis. In this case,
decreased low of Cl– to the lining of the lungs leads to an ionic imbalance that
dehydrates the mucus, which clogs the air passages.
Analysis of CFTR led to some surprises. Although CFTR is homologous
to ABC transporters, which use ATP hydrolysis to pump solutes in or out
of the cell, it shows unusual behavior. Which one of the following statements would suggest that CFTR is not a typical member of the ABC transporter family?
A. CFTR binds to Cl– to move the ions across membranes.
B. CFTR can move Cl– against a concentration gradient.
C. CFTR produces a robust Cl– current across membranes.
D. CFTR requires ATP to move C– across membranes.
11–77
he loss of luids in severe diarrhea can be efectively treated by oral
rehydration therapy, which has been called one of the greatest medical advances of the twentieth century because it dramatically reduces
deaths caused by dehydration. Oral rehydration therapy is simple and
cheap: afected individuals drink a solution of 90 mM NaCl and 110 mM
glucose. Which of the following statements best describes how oral rehydration therapy works?
A. Co-transport of Na+ and glucose via a symporter increases cell osmolarity and water uptake.
B. Glucose powers the Na+-K+ pump, which pumps Na+ into the cell to
increase cell osmolarity.
C. In the gut, the low osmolarity rehydration solution drives water into cells
lining the intestine.
D. he salt and glucose in the rehydration solution kills Vibrio cholerae, normalizing cell osmolarity.
11–78
235
Annulate Lamellae in a Human Oocyte.
These cytoplasmic whorls of double
membranes punctured by nuclear pore
complexes are found in many oocytes
and other cells, but their signiicance
and possible function(s) are still very
mysterious. Do they represent stockpiles
of nuclear membranes? Are the dark
blobs at the center stored maternal
mRNA? They are dificult to purify and
study biochemically, and compared to
other intracellular membranous structures
have received scant attention from cell
biologists.
Chapter 12
237
CHAPTER
Intracellular Compartments
and Protein Sorting
12
THE COMPARTMENTALIZATION OF CELLS
IN THIS CHAPTER
TERMS TO LEARN
cytoplasm
cytosol
gated transport
lumen
organelle
protein translocation
signal patch
signal peptidase
signal sequence
sorting signal
vesicular transport
DEFINITIONS
Match the deinition below with its term from the list above.
12–1
Contents of a cell that are contained within its plasma membrane but, in
the case of eukaryotic cells, outside the nucleus.
12–2
Protein sorting signal that consists of a speciic three-dimensional
arrangement of atoms on the folded protein’s surface.
12–3
Contents of the main compartment of the cell, excluding the nucleus and
membrane-bounded compartments such as endoplasmic reticulum and
mitochondria.
12–4
Movement of proteins through nuclear pore complexes between the
cytosol and the nucleus.
12–5
Membrane-enclosed compartment in a eukaryotic cell that has a distinct
structure, macromolecular composition, and function.
12–6
Protein sorting signal that consists of a short continuous sequence of
amino acids.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
12–7
he biological membranes that partition the cell into functionally distinct compartments are impermeable.
12–8
Like the lumen of the endoplasmic reticulum (ER), the interior of the
nucleus is topologically equivalent to the outside of the cell.
12–9
ER-bound and free ribosomes, which are structurally and functionally
identical, difer only in the proteins they happen to be making at a particular time.
12–10
Each signal sequence speciies a particular destination in the cell.
THOUGHT PROBLEMS
12–11
Discuss the following statement: “he plasma membrane is only a minor
component of most eukaryotic cells.”
THE COMPARTMENTALIZATION
OF CELLS
THE TRANSPORT OF
MOLECULES BETWEEN THE
NUCLEUS AND THE CYTOSOL
THE TRANSPORT OF PROTEINS
INTO MITOCHONDRIA AND
CHLOROPLASTS
PEROXISOMES
THE ENDOPLASMIC RETICULUM
238
Chapter 12: Intracellular Compartments and Protein Sorting
12–12
Is it really true that all human cells contain the same basic set of membrane-enclosed organelles? Do you know of any examples of human cells
that do not have a complete set of organelles?
12–13
When cells are treated with drugs that depolymerize microtubules, the
Golgi apparatus is fragmented into small vesicles and dispersed throughout each cell. When such drugs are removed, cells typically recover and
grow normally. If cells that have recovered from such treatment are
examined by electron microscopy, they are found to contain a perfectly
normal-looking Golgi apparatus. Does this mean that the Golgi apparatus has been synthesized anew from scratch? If not, how do you suppose
it might have happened?
12–14
Why do eukaryotic cells require a nucleus as a separate compartment
when prokaryotic cells manage perfectly well without?
12–15
What is the fate of a protein with no sorting signal?
12–16
Protein synthesis in a liver cell occurs nearly exclusively on free ribosomes
in the cytosol and on ribosomes that are bound to the ER membrane. (A
small fraction of total protein synthesis is directed by the mitochondrial
genome and occurs on ribosomes in the mitochondrial matrix.) Which
type of protein synthesis—in the cytosol or on the ER—do you think is
responsible for the majority of protein synthesis in a liver cell? Assume
that the average density and lifetimes of proteins are about the same in
all compartments. Explain the basis for your answer. Would your answer
change if you took into account that some proteins are secreted from
liver cells?
12–17
List the organelles in an animal cell that obtain their proteins via gated
transport, via transmembrane transport, or via vesicular transport.
Imagine that you have engineered a set of genes, each encoding a protein
with a pair of conlicting signal sequences that specify diferent compartments. If the genes were expressed in a cell, predict which signal would
win out for the following combinations. Explain your reasoning.
A. Signals for import into nucleus and import into ER.
B. Signals for import into peroxisomes and import into ER.
C. Signals for import into mitochondria and retention in ER.
D. Signals for import into nucleus and export from nucleus.
12–18
12–19
If you think of the protein as a traveler, what kind of vehicle would best
describe the sorting receptor: a private car, a taxi, or a bus? Explain your
choice.
CALCULATIONS
12–20
A typical animal cell is said to contain some 10 billion protein molecules
that need to be sorted into their proper compartments. hat’s a lot of proteins. Can 10 billion protein molecules even it into a cell? An average
protein encoded by the human genome is 450 amino acids in length, and
the average mass of an amino acid in a protein is 110 daltons. Given that
the average density of a protein is 1.4 g/cm3, what fraction of the volume
of a cell would 10 billion average protein molecules occupy? Consider a
liver cell, which has a volume of about 5000 μm3, and a pancreatic exocrine cell, which has a volume of about 1000 μm3.
12–21
he lipid bilayer, which is 5 nm thick, occupies about 60% of the volume
of a typical cell membrane. (Lipids and proteins contribute equally on a
mass basis, but lipids are less dense and therefore account for more of
the volume.) For liver cells and pancreatic exocrine cells, the total area
of all cell membranes is estimated at about 110,000 μm2 and 13,000 μm2,
THE COMPARTMENTALIZATION OF CELLS
respectively. What fraction of the total volumes of these cells is accounted
for by lipid bilayers? he volumes of liver cells and pancreatic exocrine
cells are about 5000 μm3 and 1000 μm3, respectively.
he rough ER is the site of synthesis of many classes of membrane
proteins. Some of these proteins remain in the ER, whereas others are
sorted to compartments such as the Golgi apparatus, lysosomes, and the
plasma membrane. One measure of the diiculty of the sorting problem
is the degree of “puriication” that must be achieved during transport
from the ER to the other compartments. For example, if membrane proteins bound for the plasma membrane represented 90% of all proteins in
the ER, then only a small degree of puriication would be needed (and
the sorting problem would appear relatively easy). On the other hand, if
plasma membrane proteins represented only 0.01% of the proteins in the
ER, a very large degree of puriication would be required (and the sorting
problem would appear correspondingly more diicult).
What is the magnitude of the sorting problem? What fraction of the
membrane proteins in the ER are destined for other compartments? A
few simple considerations allow one to answer these questions. Assume
that all proteins on their way to other compartments remain in the ER
for 30 minutes on average before exiting, and that the ratio of proteins to
lipids in the membranes of all compartments is the same.
A. In a typical growing cell that is dividing once every 24 hours, the equivalent of one new plasma membrane must transit the ER every day. If the
ER membrane is 20 times the area of a plasma membrane, what is the
ratio of plasma membrane proteins to other membrane proteins in the
ER?
B. If in the same cell the Golgi membrane is three times the area of the
plasma membrane, what is the ratio of Golgi membrane proteins to other
membrane proteins in the ER?
C. If the membranes of all other compartments (lysosomes, endosomes,
inner nuclear membrane, and secretory vesicles) that receive membrane
proteins from the ER are equal in total area to the area of the plasma
membrane, what fraction of the membrane proteins in the ER of this cell
are permanent residents of the ER membrane?
12–22
DATA HANDLING
12–23
Although the vast majority of transmembrane proteins insert into membranes with the help of dedicated protein-translocation machines, a few
proteins can insert into membranes on their own. Such proteins may
provide a window into how membrane insertion occurred in the days
before complex translocators had evolved.
You are studying a protein that inserts itself into the bacterial membrane independent of the normal translocation machinery. his protein
has an N-terminal, 18-amino-acid hydrophilic segment that is located on
the outside of the membrane, a 19-amino-acid hydrophobic transmembrane segment lanked by negatively and positively charged amino acids,
and a C-terminal domain that resides inside the cell (Figure 12–1A). If
the protein is properly inserted in the membrane, the N-terminal segment is exposed externally where it can be clipped of by a protease,
allowing you to quantify insertion.
To examine the roles of the hydrophobic segment and its lanking
charges, you construct a set of modiied genes that express mutant proteins with altered charges in the N-terminal segment, altered lengths of
the hydrophobic segment, and combinations of the two (Figure 12–1B).
For each gene you measure the fraction of the total protein that is cleaved
by the protease, which is the fraction that was inserted correctly (Figure
12–1B). To assess the contribution of the normal membrane potential
(positive outside, negative inside), you repeat the measurements in the
239
Chapter 12: Intracellular Compartments and Protein Sorting
OUTSIDE
INSIDE
C
(B)
percent inserted
–
–
–
3.
–
Match the deinition below with its term from the list above.
5.
–
–
Large multiprotein structure forming a channel through the nuclear
envelope that allows selected molecules to move between nucleus and
cytoplasm.
–
6.
7.
12–26
Monomeric GTPase present in both cytosol and nucleus that is required
for the active transport of macromolecules into and out of the nucleus
through nuclear pore complexes.
12–27
Fibrous meshwork of proteins on the inner surface of the inner nuclear
membrane.
12–28
Protein that binds nuclear localization signals and facilitates the transport of proteins with these signals from the cytosol into the nucleus
through nuclear pore complexes.
12–29
Sorting signal found in proteins destined for the nucleus and which enable their selective transport into the nucleus from the cytosol through the
nuclear pore complexes.
12–30
he portion of the nuclear envelope that is continuous with the endoplasmic reticulum and is studded with ribosomes on its cytosolic surface.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
12–31
he nuclear membrane is freely permeable to ions and other small molecules under 5000 daltons.
12–32
To avoid the inevitable collisions that would occur if two-way traic
through a single pore were allowed, nuclear pore complexes are specialized so that some mediate import while others mediate export.
– CCCP
+ CCCP
99
99
98
98
93
54
65
0
58
0
0
0
0
0
+
12–25
+
2.
DEFINITIONS
Sorting signal contained in the structure of macromolecules and complexes that are transported from the nucleus to the cytosol through
nuclear pore complexes.
construct
1.
4.
12–24
–
+
nuclear transport
receptor
nucleoporin
outer nuclear
membrane
Ran
+
+
nuclear lamin
nuclear lamina
nuclear localization
signal
nuclear pore complex
(NPC)
+
–
–
TERMS TO LEARN
inner nuclear membrane
nuclear envelope
nuclear export receptor
nuclear export signal
nuclear import receptor
N
–
+
THE TRANSPORT OF MOLECULES BETWEEN THE
NUCLEUS AND THE CYTOSOL
(A)
+
presence of CCCP, an ionophore that eliminates the charge on the membrane (Figure 12–1B).
A. Which of the two N-terminal negative charges is the more important for
insertion of the protein in the presence of the normal membrane potential (minus CCCP)? Explain your reasoning.
B. In the presence of the membrane potential (minus CCCP), is the hydrophobic segment or the N-terminal charge more important for insertion
of the protein into the membrane? Explain your reasoning.
C. In the absence of the membrane potential (plus CCCP), is the hydrophobic segment or the N-terminal charge more important for insertion of the
protein into the membrane? Explain your reasoning.
+
240
Figure 12–1 Insertion of a small protein into
the bacterial membrane (Problem 12–23).
(A) Normal orientation of the protein in
the membrane. (B) Mutant proteins used
to investigate the contributions of the
N-terminal negative charges and length of
the hydrophobic segment to membrane
insertion. The presence of negative charges
is indicated by –; green cylinders indicate
the length of α-helical, transmembrane
hydrophobic segments; deletions of the
hydrophobic segments are shown as gaps.
Percent inserted refers to the proportion
of the total protein whose N-termini are
sensitive to protease digestion.
THE TRANSPORT OF MOLECULES BETWEEN THE NUCLEUS AND THE CYTOSOL
nuclear
pore
CYTOSOL
nuclear
envelope
241
Figure 12–2 Cross section through a
nuclear pore complex, showing continuity
of inner and outer nuclear membranes
(Problem 12–35).
NUCLEUS
12–33
Some proteins are kept out of the nucleus, until needed, by inactivating
their nuclear localization signals by phosphorylation.
12–34
All cytosolic proteins have nuclear export signals that allow them to be
removed from the nucleus when it reassembles after mitosis.
THOUGHT PROBLEMS
12–35
As shown in Figure 12–2, the inner and outer nuclear membranes form
a continuous sheet, connecting through the nuclear pores. Continuity implies that membrane proteins can move freely between the two
nuclear membranes by difusing through the bilayer at the nuclear pores.
Yet the inner and outer nuclear membranes have diferent protein compositions, as beits their diferent functions. How do you suppose this
apparent paradox is reconciled?
12–36
How is it that a single nuclear pore complex can eiciently transport proteins that possess diferent kinds of nuclear localization signal?
12–37
How do you suppose that proteins with a nuclear export signal get into
the nucleus?
Your advisor is explaining his latest results in your weekly lab meeting.
By fusing his protein of interest to green luorescent protein (GFP), he
has shown that it is located entirely in the nucleus. But he wonders if it is
a true nuclear protein or a shuttling protein that just spends most of its
time in the nucleus. He is unsure how to resolve this issue. Having just
read an article about how a similar problem was answered, you suggest
that he make a heterokaryon by fusing cells that are expressing his tagged
protein with an excess of cells that are not expressing it. You tell him that
in the presence of a protein synthesis inhibitor to block new synthesis of
the tagged protein, he can resolve the issue by examining fused cells with
two nuclei. He gives you a puzzled look and asks, “How does that help?”
You tell him what he has so often told you: “hink about it.”
A. How would examining the two nuclei in a heterokaryon answer the question? What results would you expect if the protein were a true nuclear
protein? What would you expect if it were a shuttling protein?
B. Why did you suggest that a protein synthesis inhibitor would be needed
in this experiment?
12–38
12–39
12–40
(A)
+ cytoplasm
+ GTP
Nuclear localization signals are not cleaved of after transport into the
nucleus, whereas the signal sequences for import into other organelles
are often removed after import. Why do you suppose it is critical that
nuclear localization signals remain attached to their proteins?
To test the hypothesis that the directionality of transport across the
nuclear membrane is determined primarily by the gradient of the RanGDP outside the nucleus and Ran-GTP inside the nucleus, you decide
to reverse the gradient to see if you can force the import of a protein that
is normally exported from the nucleus. You add a well-deined nuclear
export substrate, luorescent BSA coupled with a nuclear export signal
(NES-BSA), to the standard permeabilized cell assay. Sure enough, it is
excluded from the nuclei (Figure 12–3A). Now you add Crm1, the nuclear
export receptor that recognizes the export signal, and RanQ69L-GTP, a
NES-BSA
(B)
+ Crm1
+ RanQ69L-GTP
Figure 12–3 Directionality of nuclear
transport (Problem 12–40). (A) Exclusion
of luorescent nes-Bsa from the nucleus
in a standard import assay. (B) Import of
luorescent nes-Bsa in the presence of
Crm1 and ranQ69L-GTp.
Chapter 12: Intracellular Compartments and Protein Sorting
242
mutant form of Ran that cannot hydrolyze GTP. With these additions, the
tagged BSA now enters the nuclei (Figure 12–3B). Unlike conventional
nuclear import, which concentrates imported proteins in the nucleus,
the concentration of NES-BSA in the nucleus in the import assay is no
higher than in the surrounding cytoplasm.
A. Why doesn’t NES-BSA accumulate to a higher concentration in the
nucleus than in the cytoplasm in these experiments?
B. In a standard nuclear import assay with added cytoplasm and GTP, proteins with a nuclear localization signal accumulate to essentially 100% in
the nucleus. How is it that the standard assay allows 100% accumulation
in the nucleus?
CALCULATIONS
12–41
By following the increase in nuclear luorescence over time in the forced
nuclear import experiments in Figure 12–3B, the authors were able to
show that nuclear luorescence reached half its maximal value in 60 seconds. Since the added concentration of NES-BSA was 0.3 μM, its concentration in the nucleus after 60 seconds was 0.15 μM. Given that the volume of the nucleus is 500 fL and that each nucleus contains 3000 nuclear
pores, calculate the rate of import of NES-BSA per pore in this experiment. (For this calculation, neglect export of NES-BSA.) Is your answer
physiologically reasonable? Why or why not?
12–42
A nuclear pore can dilate to accommodate a gold particle 26 nm in diameter. If it could accommodate a spherical protein of the same dimensions,
what would the protein’s molecular mass (g/mole) be? [Assume that the
density of the protein is 1.4 g/cm3. he volume of a sphere is (4/3)πr3.]
12–43
Assuming that 32 million histone octamers are required to package the
human genome, how many histone molecules must be transported per
second per nuclear pore complex in cells whose nuclei contain 3000
nuclear pores and are dividing once per day?
12–44
he nuclear pore complex (NPC) creates a barrier to the free exchange of
molecules between the nucleus and cytoplasm, but in a way that remains
mysterious. In yeast, for example, the central pore has a diameter of 35
nm and is 30 nm long, which is somewhat smaller than its vertebrate
counterpart. Even so, it is large enough to accommodate virtually all
components of the cytosol. Yet the pore allows passive difusion of molecules only up to about 40 kd; entry of anything larger requires help from
a nuclear import receptor. Selective permeability is controlled by protein
components of the NPC that have unstructured, polar tails extending
into the central pore. hese tails are characterized by periodic repeats of
the hydrophobic amino acids phenylalanine (F) and glycine (G).
At high enough concentration (50 mM), the FG-repeat domains of
these proteins can form a gel, with a meshwork of interactions between
the hydrophobic FG repeats (Figure 12–4A). hese gels allow passive diffusion of small molecules, but they prevent entry of larger proteins such
as the luorescent protein mCherry fused to maltose binding protein
(A)
(B)
solution
gel
30 s
10 min
30 min
Figure 12–4 FG-repeat gel and inlux of proteins into the nucleus (Problem
12–44). (a) Cartoon of the meshwork (gel) formed by pairwise interactions
between hydrophobic FG repeats. For FG-repeats separated by 17 amino
acids, as is typical, the mesh formed by extended amino acid side chains
would correspond to about 4 nm on a side, which would be large enough
to account for the characteristic passive diffusion of proteins through
nuclear pores. (B) Diffusion of MBp-mCherry and importin-MBp-GFp
into a gel of FG-repeats. In each group, the solution is shown at left
and the gel at right. The bright areas indicate regions that contain the
luorescent proteins.
MBP-mCherry
30 s
10 min
30 min
importin-MBP-GFP
THE TRANSPORT OF MOLECULES BETWEEN THE NUCLEUS AND THE CYTOSOL
(MBP) (Figure 12–4B). (he fusion to MBP makes mCherry too large to
enter the nucleus by passive difusion.) However, if the nuclear import
receptor, importin, is fused to a similar protein, MBP-GFP, the importinMBP-GFP fusion readily enters the gel (Figure 12–4B).
A. FG-repeats only form gels in vitro at relatively high concentration
(50 mM). Is this concentration reasonable for FG repeats in the NPC
core? In yeast, there are about 5000 FG-repeats in each NPC. Given the
dimensions of the yeast nuclear pore (35 nm diameter and 30 nm length),
calculate the concentration of FG-repeats in the cylindrical volume of the
pore. Is this concentration comparable to the one used in vitro?
B. Is difusion of importin-MBP-GFP through the FG-repeat gel fast enough
to account for the eicient low of materials between the nucleus and
cytosol? From experiments of the type shown in Figure 12–4B, the diffusion coeicient (D) of importin-MBP-GFP through the FG-repeat gel
was determined to be about 0.1 μm2/sec. he equation for difusion is
t = x2/2D, where t is time and x is distance. Calculate the time it would
take importin-MBP-GFP to difuse through a yeast nuclear pore (30 nm)
if the pore consisted of a gel of FG-repeats. Does this time seem fast
enough for the needs of a eukaryotic cell?
243
nucleoplasmin
preparation
nuclear
injection
cytoplasmic
injection
intact
one tail
heads only
tails only
DATA HANDLING
Before nuclear pore complexes were well understood, it was unclear
whether nuclear proteins difused passively into the nucleus and accumulated there by binding to residents of the nucleus such as chromosomes, or whether they were actively imported and accumulated regardless of their ainity for nuclear components.
A classic experiment that addressed this problem used several forms
of radioactive nucleoplasmin, which is a large pentameric protein
involved in chromatin assembly. In this experiment, either the intact
protein or the nucleoplasmin heads, tails, or heads with a single tail were
injected into the cytoplasm of a frog oocyte or into the nucleus (Figure
12–5). All forms of nucleoplasmin, except heads, accumulated in the
nucleus when injected into the cytoplasm, and all forms were retained in
the nucleus when injected there.
A. What portion of the nucleoplasmin molecule is responsible for localization in the nucleus?
B. How do these experiments distinguish between active transport, in
which a nuclear localization signal triggers transport by the nuclear pore
complex, and passive difusion, in which a binding site for a nuclear
component allows accumulation in the nucleus?
12–45
12–46
You have just joined a laboratory that is analyzing the nuclear transport
machinery in yeast. Your advisor, who is known for her extraordinarily
clever ideas, has given you a project with enormous potential. In principle, it would allow a genetic selection for conditional-lethal mutants in
the nuclear transport apparatus.
She gave you two plasmids. Each plasmid consists of a hybrid gene
under the control of a regulatable promoter (Figure 12–6). he hybrid
gene is a fusion between a gene whose product is normally imported into
the nucleus and the gene for the restriction nuclease EcoRI. he plasmid
pNL+ contains a functional nuclear localization signal (NLS); the plasmid
pNL– does not have an NLS. he promoter, which is from the yeast Gal1
gene, allows transcription of the hybrid gene only when the sugar galactose is present in the growth medium.
Following her instructions, you introduce the plasmids into yeast
(in the absence of galactose) and then assay the transformed yeast in
medium containing glucose and in medium containing galactose. Your
results are shown in Table 12–1. You don’t remember what your advisor
Figure 12–5 Cellular location of injected
nucleoplasmin and nucleoplasmin
components (Problem 12–45). Schematic
diagrams of autoradiographs show the
cytoplasm and nucleus with the location
of nucleoplasmin indicated by the red
areas.
pNL+
nuclear protein
Gal1
promoter
EcoRI
NLS
present
pNL–
nuclear protein EcoRI
Gal1
promoter
NLS
absent
Figure 12–6 Two plasmids for
investigating nuclear localization in yeast
(Problem 12–46). Plasmids are shown
as linear molecules for clarity. Arrows
indicate direction of transcription from the
Gal1 promoter.
Chapter 12: Intracellular Compartments and Protein Sorting
244
TABLE 12–1 Results of proliferation experiments with yeast carrying
plasmids pNL+ or pNL– (Problem 12–46).
(A)
+ GTP
– GTP
Ran
Plasmid
Glucose medium
Galactose medium
pNL+
proliferation
death
pNL–
proliferation
proliferation
told you to expect, but you know you will be expected to explain these
results at the weekly lab meeting.
Why do yeasts with the pNL+ plasmid proliferate in the presence of
glucose but die in the presence of galactose, whereas yeasts with the
pNL– plasmid proliferate in both media?
How might you use this system for a selection assay to isolate cells
defective in nuclear transport?
Puriied importin in the presence of Ran and GTP promotes uptake of
a labeled substrate into nuclei (Figure 12–7A). No uptake occurs in the
absence of GTP, and Ran alone is unable to promote nuclear uptake.
Importin by itself causes a GTP-independent accumulation of substrate
at the nuclear periphery, but does not promote nuclear uptake (Figure
12–7A).
To deine the steps in the uptake pathway, you irst incubate nuclei
with substrate in the presence of importin. You then wash away free
importin and substrate and incubate a second time with Ran and GTP
(Figure 12–7B).
A. Why do you think the substrate accumulates at the nuclear periphery, as
is seen in the absence of GTP or with importin alone in the presence of
GTP?
B. To the extent these data allow, deine the order of events that leads to
uptake of substrate into the nucleus.
Ran
+ importin
importin
(B)
1st incubation:
2nd incubation:
importin + substrate
Ran + GTP
12–47
12–48
he structures of Ran-GDP and Ran-GTP (actually Ran-GppNp, a stable
GTP analog) are strikingly diferent, as shown in Figure 12–8. Not surprisingly, Ran-GDP binds to a diferent set of proteins than does RanGTP.
To look at the uptake of Ran itself into the nuclei of permeabilized
cells, you attach a red luorescent tag to a cysteine side chain in Ran
to make it visible. his modiied Ran supports normal nuclear uptake.
Fluorescent Ran-GDP is taken up by nuclei only if cytoplasm is added,
whereas a mutant form, RanQ69L-GTP, which is unable to hydrolyze
GTP, is not taken up in the presence or absence of cytoplasm. To identify
the cytoplasmic protein that is crucial for Ran-GDP uptake, you construct
ainity columns with bound Ran-GDP or bound RanQ69L-GTP and pass
Ran-GppNp
Ran-GDP
Figure 12–8 structures of ran-GDp and ran-Gppnp (Problem 12–48). The segment
of ran shown in red displays two dramatically different conformations depending on
whether GDp or the GTp analog is bound.
Figure 12–7 Effects of Ran, importin, and
GTP on nuclear uptake of luoresceinlabeled substrate (Problem 12–47).
(a) Comparison of various combinations
of ran and importin in the presence and
absence of GTp. (B) Two-stage incubation
of cell remnants with importin and
substrate and then with ran and GTp.
Circles are the nuclei; very light circles are
nuclei without bound substrate.
THE TRANSPORT OF MOLECULES BETWEEN THE NUCLEUS AND THE CYTOSOL
12–49
he broad-spectrum antibiotic leptomycin B inhibits nuclear export, but
how does it work? In the yeast S. pombe, resistance to leptomycin B can
arise by mutations in the Crm1 gene, which encodes a nuclear export
receptor for proteins with leucine-rich nuclear export signals. To look at
nuclear export directly, you modify the green luorescent protein (GFP)
by adding a nuclear export signal (NES). In both wild-type and mutant
cells that are resistant to leptomycin B, NES-GFP is found exclusively
in the cytoplasm in the absence of leptomycin B (Figure 12–10). In the
presence of leptomycin B, however, NES-GFP is present in the nuclei of
wild-type cells, but in the cytoplasm of mutant cells (Figure 12–10). Is this
result the one you would expect if leptomycin B blocked nuclear export?
Why or why not?
Frog oocytes are a useful experimental system for studying nuclear
export because they are large cells with large nuclei. It is easy (with practice) to inject oocytes with labeled RNA and to separate the nucleus and
cytoplasm to follow the fate of the injected label. You inject a mixture
of various 32P-labeled RNA molecules into the nucleus in the presence
and absence of leptomycin B to study its efect on nuclear export of RNA.
Immediately after injection and three hours later, you analyze total (T),
cytoplasmic (C), and nuclear (N) contents by polyacrylamide-gel electrophoresis and autoradiography (Figure 12–11).
A. How good was your injection technique? Did you actually inject into the
nucleus? Did you rip apart the nuclear envelope when you injected the
RNAs? How do you know?
B. Which, if any, of the RNAs are normally exported from the nucleus?
C. Is the export of any of the RNAs inhibited by leptomycin B? What does
your answer imply about export of this collection of RNAs?
Ra
nG
D
P
Ra
nQ
69
LG
TP
cytoplasm through them. Cytoplasm passed over a Ran-GDP column no
longer supports nuclear uptake of Ran-GDP, whereas cytoplasm passed
over a column of RanQ69L-GTP retains this activity. You elute the bound
proteins from each column and analyze them on an SDS polyacrylamide
gel, looking for diferences that might identify the factor that is required
for nuclear uptake of Ran (Figure 12–9).
A. Why did you use RanQ69L-GTP instead of Ran-GTP in these experiments? Could you have used Ran-GppNp instead of RanQ69L-GTP to
achieve the same purpose?
B. Which of the many proteins eluted from the two diferent ainity columns is a likely candidate for the factor that promotes nuclear import of
Ran-GDP?
C. What other protein or proteins would you predict the Ran-GDP import
factor would bind in order to carry out its function?
D. How might you conirm that the factor you have identiied is necessary
for promoting the nuclear uptake of Ran?
245
kd
158
116
97
68
46
27
20
14
7
1
2
Figure 12–9 Proteins eluted from RanGDP (lane 1) and RanQ69l-GTP (lane
2) afinity columns (Problem 12–48). The
molecular masses of marker proteins are
shown on the left.
12–50
leptomycin B, nM
time, hr
0
0
100
0
3
3
T CN T CN T CN
mRNA
– leptomycin B
NES-GFP
+ leptomycin B
NES-GFP
U1 snRNA
U5 snRNA
U6 snRNA
DNA
DNA
tRNA
wild type
mutant
wild type
mutant
Figure 12–10 Distribution of nes-GFp in S. pombe in the presence and absence of
leptomycin B (Problem 12–49). Light areas in the nes-GFp panels show the position of
GFp. Light areas in the Dna panels result from a stain that binds to Dna and marks the
position of the nuclei in the cells in the nes-GFp panels.
Figure 12–11 effects of leptomycin B
on export of various rnas from frog
oocyte nuclei (Problem 12–50). Total (T),
cytoplasmic (C), and nuclear (n) fractions
are indicated.
246
Chapter 12: Intracellular Compartments and Protein Sorting
THE TRANSPORT OF PROTEINS INTO
MITOCHONDRIA AND CHLOROPLASTS
TERMS TO LEARN
chloroplast
inner membrane
intermembrane space
matrix space
mitochondria
mitochondrial hsp70
mitochondrial precursor
protein
outer membrane
OXA complex
porin
protein translocator
SAM complex
stroma
thylakoid
TIM complex
TOM complex
DEFINITIONS
Match the deinition below with its term from the list above.
12–51
Membrane-enclosed organelles, about the size of bacteria, that carry out
oxidative phosphorylation and produce most of the ATP in eukaryotic
cells.
12–52
Part of a multisubunit protein assembly that is bound to the matrix side
of the TIM23 complex and acts as a motor to pull the precursor protein
into the matrix space.
12–53
Multisubunit protein assembly that transports proteins across the mitochondrial outer membrane.
12–54
he matrix space of a chloroplast.
12–55
he membrane of a mitochondrion that encloses the matrix and is folded
into cristae.
12–56
Central subcompartment of a mitochondrion, enclosed by the inner
mitochondrial membrane.
12–57
Flattened sac of membrane in a chloroplast that contains the protein
subunits of the photosynthetic system and of the ATP synthase.
12–58
Protein encoded by a nuclear gene, synthesized in the cytosol, and subsequently transported into mitochondria.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
12–59
he TOM complex is required for the import of all nucleus-encoded
mitochondrial proteins.
12–60
he two signal sequences required for transport of nucleus-encoded proteins into the mitochondrial inner membrane via the TIM23 complex are
cleaved of the protein in diferent mitochondrial compartments.
12–61
Import of proteins into mitochondria and chloroplasts is very similar;
even the individual components of their transport machinery are homologous, as beits their common evolutionary origin.
THOUGHT PROBLEMS
12–62
To aid your studies of protein import into mitochondria, you treat yeast
cells with cycloheximide, which blocks ribosome movement along
mRNA. When you examine these cells in the electron microscope, you
are surprised to ind cytosolic ribosomes attached to the outside of the
mitochondria. You have never seen attached ribosomes in the absence
THE TRANSPORT OF PROTEINS INTO MITOCHONDRIA AND CHLOROPLASTS
of cycloheximide. To investigate this phenomenon further, you prepare
mitochondria from cells that have been treated with cycloheximide and
then extract the mRNA that is bound to the ribosomes associated with
the mitochondria. You translate this mRNA in vitro and compare the
protein products with similarly translated mRNA from the cytosol. he
results are clear-cut: the mitochondria-associated ribosomes are translating mRNAs that encode mitochondrial proteins.
You are astounded! Here, clearly visible in the electron micrographs,
seems to be proof that protein import into mitochondria occurs during
translation. How might you reconcile this result with the prevailing view
that mitochondrial proteins are imported only after they have been synthesized and released from ribosomes?
12–63
You have made a peptide that contains a functional mitochondrial import
signal. Would you expect the addition of an excess of this peptide to afect
the import of mitochondrial proteins? Why or why not?
12–64
Components of the TIM complexes, the multisubunit protein translocators in the mitochondrial inner membrane, are much less abundant
than those of the TOM complex. hey were initially identiied using
a genetic trick. he yeast Ura3 gene, whose product is an enzyme that
is normally located in the cytosol where it is essential for synthesis of
uracil, was modiied so that the protein carried an import signal for the
mitochondrial matrix. A population of cells carrying the modiied Ura3
gene in place of the normal gene was then grown in the absence of uracil. Most cells died, but the rare cells that grew were shown to be defective for import into the mitochondrial matrix. Explain how this selection
identiies cells with defects in components required for import into the
mitochondrial matrix. Why don’t normal cells with the modiied Ura3
gene grow in the absence of uracil? Why do cells that are defective for
mitochondrial import grow in the absence of uracil?
12–65
Mitochondria normally provide cells with most of the ATP they require
to meet their energy needs. Mitochondria that cannot import proteins
are defective for ATP synthesis. How is it that cells with import-defective
mitochondria can survive at all? How do they get the ATP they need to
function?
12–66
Describe in a general way how you might use radiolabeled proteins and
proteases to study import processes in isolated, intact mitochondria.
What sorts of experimental controls might you include to ensure that the
results you obtain mean what you think they do?
12–67
If the enzyme dihydrofolate reductase (DHFR), which is normally located
in the cytosol, is engineered to carry a mitochondrial targeting sequence
at its N-terminus, it is eiciently imported into mitochondria. If the modiied DHFR is irst incubated with methotrexate, which binds tightly to
the active site, the enzyme remains in the cytosol. How do you suppose
that the binding of methotrexate interferes with mitochondrial import?
12–68
Why do mitochondria need a special translocator to import proteins
across the outer membrane, when the membrane has already has large
pores formed by porins?
CALCULATIONS
12–69
he vast majority of mitochondrial proteins are imported through the
outer membrane by the multisubunit protein translocators known as
TOM complexes. Since the number of TOM complexes in yeast mitochondria is known (10 pmole/mg of mitochondrial protein), it is possible to calculate whether a co-translational mechanism could account
247
248
Chapter 12: Intracellular Compartments and Protein Sorting
for the bulk of mitochondrial protein import. If mitochondrial proteins
were all imported co-translationally, then 10 pmol of TOM complexes
would need to import 1 mg of protein each generation. Given that mitochondria double every 3 hours and that the rate of protein synthesis is 3
amino acids per second (which is therefore the maximum rate of import
through a single TOM complex), how many milligrams of mitochondrial
protein could 10 pmol of TOM complexes import in one generation? (On
average, an amino acid has a mass of 110 daltons.)
DATA HANDLING
12–70
Barnase is a 110-amino-acid bacterial ribonuclease that is often used
as a model for studies of protein folding and unfolding. It forms a compact folded structure that has a high energy of activation for unfolding
(about 85 kJ/mole). Can such a protein be imported into mitochondria?
To the N-terminus of barnase, you add 35, 65, or 95 amino acids from the
N-terminus of pre-cytochrome b2, all of which include the cytochrome’s
mitochondrial import signal. N35-barnase is not imported, N65-barnase
is imported at a low rate, and N95-barnase is imported very eiciently
into isolated mitochondria. None of these N-terminal extensions have
any measurable efect on the stability of the barnase domain. If these
proteins are denatured before testing for import, they are all imported at
the same high rate. How do you suppose that longer N-terminal extensions facilitate the import of barnase?
12–71
Are proteins imported into mitochondria as completely unfolded polypeptide chains, or can the translocation apparatus accommodate fully
or partially folded structures? hat is, is the protein sucked up like a noodle, or is it swallowed whole, as a python devours its prey? It is possible to engineer cysteine amino acids into barnase and then cross-link
them to make disulide bonds either between C5 and C78 or between
C43 and C80 (Figure 12–12). Import of N95-barnase (see Problem 12–70)
was tested in the presence and absence of disulide cross-links at these
two positions. Its import was unafected by either cross-link. By contrast,
import of N65-barnase was blocked by the C5–C78 cross-link but unaffected by the C43–C80 cross-link. Do these results allow you to distinguish between import of extended polypeptide chains or of folded structures? Why or why not?
12–72
Tim23 is a key component of the mitochondrial TIM23 protein translocator complex. he N-terminal half of Tim23 is hydrophilic, while the C-terminal half is hydrophobic and probably spans the membrane four times,
as suggested by the hydropathy plot in Figure 12–13A. To determine the
arrangement of Tim23 in mitochondrial membranes, a protease was
added to intact mitochondria or to mitoplasts, which are mitochondria
barnase (C5–C78)
S
N-terminus
S
barnase (C43–C80)
C-terminus
S
S
N-terminus
C-terminus
Figure 12–12 Structures of barnase
molecules with disulide bonds
between cysteines at positions 5 and
78 (C5–C78) or between positions 43
and 80 (C43–C80) (Problem 12–71).
PEROXISOMES
249
(A)
(B)
hydrophobic
mitochondria
mitoplasts
protease
+
–
–
+
–
+
1
2
Figure 12–13 Arrangement of Tim23
in mitochondrial membranes (Problem
12–72). (A) Hydropathy plot for Tim23.
(B) Sensitivity of Tim23 in mitochondria
and mitoplasts to digestion with a
protease.
–
+
+
hydrophilic
50
100
150
residue number
200
antibodies
specific for
C-terminus
antibodies
specific for
N-terminal
half
3
from which the outer membranes have been removed. he mobility of
Tim23 was detected on SDS polyacrylamide gels by immunoblotting
with antibodies speciic for the N-terminal half or for the extreme C-terminus of Tim23 (Figure 12–13B). Normal-sized Tim23 was present in
both mitochondria and mitoplasts (as shown for mitochondria in Figure
12–13B), but Tim23 was partially digested when mitochondria and mitoplasts were treated with a protease (Figure 12–13B).
A. Is Tim23 an integral component of the inner or outer mitochondrial
membrane? Explain your reasoning.
B. To the extent the information in this problem allows, diagram the
arrangement of Tim23 in mitochondrial membranes.
PEROXISOMES
TERMS TO LEARN
peroxin
peroxisome
DEFINITIONS
Match the deinition below with its term from the list above.
12–73
Small membrane-bounded organelle that uses molecular oxygen to oxidize organic molecules.
12–74
One of several proteins that are involved in protein import into peroxisomes.
TRUE/FALSE
Decide whether this statement is true or false, and then explain why.
12–75
Peroxisomes are found in only a few specialized types of eukaryotic cell.
THOUGHT PROBLEMS
12–76
Catalase, an enzyme normally found in peroxisomes, is present in normal amounts in cells that do not have visible peroxisomes. It is possible
to determine the location of catalase in such cells using immunoluorescence microscopy with antibodies speciic for catalase. Fluorescence
micrographs of normal cells and peroxisome-deicient cells are shown
in Figure 12–14. Where is catalase located in cells without peroxisomes
(Figure 12–14B)? Why does catalase show up as small dots of luorescence in normal cells (Figure 12–14A)?
Chapter 12: Intracellular Compartments and Protein Sorting
250
(A)
Figure 12–14 location of catalase in cells
as determined by immunoluorescence
microscopy (Problem 12–76). (a) normal
cells. (B) peroxisome-deicient cells. Cells
were reacted with antibodies speciic for
catalase, washed, and then stained with
a luorescein-labeled second antibody
that is speciic for the catalase-speciic
antibody. The two panels are at the
same magniication; a 10 μm scale bar is
shown in (B).
(B)
10 µm
12–77
Cells with functional peroxisomes incorporate 9-(1ʹ-pyrene)nonanol
(P9OH) into membrane lipids. Exposure of such cells to ultraviolet (UV)
light causes cell death because excitation of the pyrene moiety generates
reactive oxygen species, which are toxic to cells. Cells that do not make
peroxisomes lack a critical enzyme responsible for incorporating P9OH
into membrane lipids. How might you make use of P9OH to select for
cells that are missing peroxisomes?
DATA HANDLING
You have isolated two mutant cell lines that lack typical peroxisomes.
When you test these cells for peroxisomal enzymes, you ind that catalase activity is virtually the same as in normal cells. By contrast, acyl
CoA oxidase activity is absent in both mutant cell lines. To investigate
the acyl CoA oxidase deiciency, you perform a pulse-chase experiment:
you grow cells for 1 hour in medium containing 35S-methionine, then
transfer them to unlabeled medium and immunoprecipitate acyl CoA
oxidase at various times after transfer (Figure 12–15). You observe two
forms of oxidase in normal cells, but only one in the mutant cell lines. To
clarify the relationship between the 75 kd and 53 kd forms of the oxidase,
you isolate mRNA from wild-type and the mutant cell lines, translate it
in vitro, and immunoprecipitate acyl CoA oxidase. All three sources of
mRNA give similar levels of the 75 kd form, but none of the 53 kd form.
A. How do you think the two forms of acyl CoA oxidase in normal cells are
related? Which one, if either, do you suppose is the active enzyme?
B. Why do the mutant cells have only the 75 kd form of acyl CoA oxidase,
and why do you think it disappears during the pulse-chase experiment?
If you had done a similar experiment with catalase, do you suppose it
would have behaved the same way?
12–78
12–79
Proteins that are imported into the peroxisome matrix using a C-terminal tripeptide signal are recognized by the cytosolic receptor Pex5, which
docks at a complex of proteins in the peroxisomal membrane. Delivery by
a cytosolic receptor distinguishes peroxisomal import from import into
mitochondria, chloroplasts, and the endoplasmic reticulum. Moreover,
unlike import into those organelles, peroxisomes can import fully folded
proteins and protein oligomers. You wish to distinguish three potential
modes of action for Pex5: (1) it delivers its cargo to the peroxisomal membrane but remains cytosolic; (2) it enters the peroxisome along with its
cargo; or (3) it cycles between the cytosol and the peroxisomal matrix.
normal CHO cells
chase (hours)
0
1
3
8
24
mutant 1
0
1
3
8
mutant 2
24
0
1
3
8
24
75 kd
53 kd
Figure 12–15 pulse-chase experiments
with normal Chinese hamster ovary (ChO)
cells and mutant cells (Problem 12–78).
PEROXISOMES
251
Figure 12–16 Mechanism of Pex5-mediated peroxisomal import (Problem
12–79). (A) Modiied pex5. (B) expectations for different mechanisms of
pex5-mediated peroxisomal import. (C) analysis of transfection of the
modiied Pex5 gene into cells. “WCe” stands for whole-cell extract, “p” for
pellet, and “s” for supernatant. proteins were detected by immunoblotting
using mab1 or mab2, followed by reaction with a second antibody that
binds mab1 and mab2 and also carries bound horseradish peroxidase,
which then converts an added precursor into a light-emitting molecule that
can be detected on photographic ilm. Black bands correspond to sites
where the ilm has been exposed by the light-emitting molecule. In (B),
uncleaved pex5 (upper band) and cleaved pex5 (lower band) are separated
by a larger distance than they are in the experiment (C) for clarity.
(A) CONSTRUCT
cleavage
site
FLAG tag
Pex5
peptide
(B) THEORETICAL
WCE P
S
WCE P
S
cytosolic
To deine the mechanism for Pex5-mediated import, you modify
the Pex5 gene to encode an N-terminal peptide segment that includes a
cleavage site for a protease localized exclusively to the matrix of the peroxisome (Figure 12–16A). Immediately adjacent to the cleavage site is a
sequence of amino acids (the so-called FLAG tag) that can be recognized
by commercial antibodies. One antibody, mAb2, binds the FLAG tag in
any context, whereas another, mAb1, binds the FLAG tag only when it is
at the N-terminus and thus will detect only cleaved Pex5. By preparing a
whole-cell extract (WCE) and fractionating it into a pellet (P), which contains the peroxisomes, and supernatant (S), which contains the cytosol,
you can distinguish the three possible mechanisms of Pex5-mediated
import—cytosolic, imported, cycling—by using the mAb1 and mAb2
antibodies, as shown in Figure 12–16B.
When you express the modiied Pex5 gene in cells, prepare cell fractions, separate the proteins by electrophoresis, and react them with mAb1
and mAb2 antibodies, you obtain the results shown in Figure 12–16C.
A. Explain the theoretical results (Figure 12–16B) expected for each of the
three possible mechanisms of Pex5-mediated import.
B. Based on the results in Figure 12–16C, how does Pex5 mediate the import
of proteins into the matrix of the peroxisome? Explain your reasoning.
C. Pex5-mediated import into peroxisomes resembles most closely import
into what other cellular organelle?
MEDICAL LINKS
12–80
Primary hyperoxaluria type 1 (PH1) is a lethal autosomal recessive disease caused by a deiciency of the liver-speciic peroxisomal enzyme
alanine:glyoxylate aminotransferase (AGT). About one-third of PH1
patients possess signiicant levels of AGT protein and enzyme activity.
Analysis of these patients shows that their AGT contains two critical single amino acid changes: one that interferes with peroxisomal targeting
and a second that allows the N-terminus to form an amphiphilic α helix
with a positively charged side. Where do you suppose this mutant AGT is
found in cells from these patients? Explain your reasoning.
12–81
Trypanosomes are single-celled parasites that cause sleeping sickness when they infect humans. Trypanosomes from humans carry the
enzymes for a portion of the glycolytic pathway in a peroxisomelike
organelle, termed the glycosome. By contrast, trypanosomes from the
tsetse ly—the intermediate host—carry out glycolysis entirely in the
cytosol. his intriguing diference has alerted the interest of the pharmaceutical company that employs you. Your company wishes to exploit this
diference to control the disease.
You decide to study the enzyme phosphoglycerate kinase (PGK)
because it is in the afected portion of the glycolytic pathway. Trypanosomes from the tsetse ly express PGK entirely in the cytosol, whereas
trypanosomes from humans express 90% of the total PGK activity in glycosomes and only 10% in the cytosol. When you clone PGK genes from
imported
cycling
mAb2
mAb1
(C) EXPERIMENTAL
WCE P
mAb2
S
WCE P
mAb1
S
Chapter 12: Intracellular Compartments and Protein Sorting
252
trypanosomes, you ind three forms that difer slightly from one another.
Exploiting these small diferences, you design three oligonucleotides
that hybridize speciically to the mRNAs from each gene. Using these oligonucleotides as probes, you determine which genes are expressed by
trypanosomes from humans and from tsetse lies. he results are shown
in Figure 12–17.
A. Which PGK genes are expressed in trypanosomes from humans? Which
are expressed in trypanosomes from tsetse lies?
B. Which PGK gene probably encodes the glycosomal form of PGK?
C. Do you think that the minor cytosolic PGK activity in trypanosomes
from humans is due to inaccurate sorting into glycosomes? Explain your
answer.
THE ENDOPLASMIC RETICULUM
TERMS TO LEARN
BiP
calnexin
calreticulin
co-translational
dolichol
endoplasmic reticulum (ER)
ER lumen
ER resident protein
ER retention signal
ER signal sequence
ER tail-anchored protein
free ribosome
glycoprotein
GPI anchor
membrane-bound ribosome
microsome
multipass transmembrane protein
polyribosome
post-translational
protein glycosylation
rough ER
Sec61 complex
signal-recognition particle (SRP)
single-pass transmembrane protein
smooth ER
SRP receptor
start-transfer signal
stop-transfer signal
translocon
unfolded protein response
DEFINITIONS
Match the deinition below with its term from the list above.
12–82
Region of the ER not associated with ribosomes.
12–83
Type of lipid linkage by which some proteins are bound to the membrane.
12–84
Labyrinthine membrane-enclosed compartment in the cytoplasm of
eukaryotic cells, where lipids are synthesized and membrane-bound
proteins and secretory proteins are made.
12–85
Ribonucleoprotein complex that binds an ER signal sequence on a partially synthesized polypeptide chain and directs the polypeptide and its
attached ribosome to the ER.
12–86
Hydrophobic amino acid sequence that halts translocation of a polypeptide chain through the ER membrane, thus anchoring the protein chain
in the membrane.
12–87
Describes import of a protein into the ER before the polypeptide chain is
completely synthesized.
12–88
Short amino acid sequence on a protein that keeps it in the ER.
12–89
Cellular action triggered by an accumulation of misfolded proteins in the
ER.
12–90
he protein translocator that forms a water-illed pore in the ER
probe
source
of mRNA
oligo 1
oligo 2
oligo 3
H
H
H
F
F
F
gene 3
gene 1
gene 2
Figure 12–17 Hybridization of speciic
oligonucleotide probes (oligos) to
mrna isolated from trypanosomes
from humans (h) and tsetse lies (F)
(Problem 12–81). The intensity of the
bands on the autoradiograph relects
the concentrations of the mrnas.
THE ENDOPLASMIC RETICULUM
253
membrane, allowing passage of a polypeptide chain as it is being synthesized by membrane-bound ribosomes.
12–91
Any protein with one or more oligosaccharide chains covalently linked to
amino acid side chains.
12–92
Ribosome in the cytosol that is unattached to any membrane.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
12–93
he signal peptide binds to a hydrophobic site on the ribosome causing a
pause in protein synthesis, which resumes when SRP binds to the signal
peptide.
12–94
Nascent polypeptide chains are transferred across the ER membrane
through a pore in the Sec61 protein translocator complex.
12–95
In multipass transmembrane proteins, the odd-numbered transmembrane segments (counting from the N-terminus) act as start-transfer signals and the even-numbered segments act as stop-transfer signals.
12–96
he ER lumen contains a mixture of thiol-containing reducing agents
that prevent the formation of S–S linkages (disulide bonds) by maintaining the cysteine side chains of luminal proteins in reduced (–SH) form.
THOUGHT PROBLEMS
12–97
Explain how an mRNA molecule can remain attached to the ER membrane while the individual ribosomes translating it are released and
rejoin the cytosolic pool of ribosomes after each round of translation.
12–98
Why are cytosolic hsp70 chaperone proteins required for import of proteins into mitochondria and chloroplasts, but not for co-translational
import into the ER?
12–99
Where would you expect to see microsomes in an electron micrograph of
a liver cell?
12–100 Compare and contrast protein import into the ER and into the nucleus.
List at least two major diferences in the mechanisms and speculate on
why the nuclear mechanism might not work for ER import and vice versa.
12–101 Four membrane proteins are represented schematically in Figure 12–18.
he boxes represent membrane-spanning segments and the arrows represent sites for cleavage of the signal sequence. Predict how each of the
mature proteins will be arranged across the membrane of the ER. Indicate
(A)
N
1
2
(B)
+
–
1
N
(C)
C
–
+
1
N
C
2
3
4
C
(D)
N
1N
2
3
4
5
C
Figure 12–18 Distribution of the
membrane-spanning segments in
proteins to be inserted into the ER
membrane (Problem 12–101). Boxes
represent membrane-spanning segments
and arrows indicate sites at which signal
sequences are cleaved. The pluses and
minuses indicate the charged amino acids
at the ends of some transmembrane
segments.
254
Chapter 12: Intracellular Compartments and Protein Sorting
COOH
1
3
5
CYTOSOL
ER LUMEN
2
4
6
NH2
clearly the N- and C-termini relative to the cytosol and the lumen of the
ER, and label each box as a start-transfer or stop-transfer signal.
12–102 Examine the multipass transmembrane protein shown in Figure 12–19.
What would you predict would be the efect of converting the irst hydrophobic transmembrane segment to a hydrophilic segment? Sketch the
arrangement of the modiied protein in the ER membrane.
12–103 Why might it be advantageous to add a preassembled block of 14 sugars
to a protein in the ER, rather than building the sugar chains step-by-step
on the surface of the protein by the sequential addition of sugars by individual enzymes?
12–104 Outline the steps by which misfolded proteins in the ER trigger synthesis
of additional ER chaperone proteins. How does this response beneit the
cell?
12–105 All new phospholipids are added to the cytosolic lealet of the ER mem-
brane, yet the ER membrane has a symmetrical distribution of diferent
phospholipids in its two lealets. By contrast, the plasma membrane,
which receives all its membrane components ultimately from the ER, has
a very asymmetrical distribution of phospholipids in the two lealets of
its lipid bilayer. How is the symmetry generated in the ER membrane,
and how is the asymmetry generated and maintained in the plasma
membrane?
DATA HANDLING
12–106 Translocation of proteins across rough microsomal membranes can be
judged by several experimental criteria: (1) the newly synthesized proteins are protected from added proteases, unless detergents are present
to solubilize the lipid bilayer; (2) the newly synthesized proteins are glycosylated by oligosaccharide transferases, which are localized exclusively
to the lumen of the ER; (3) the signal peptides are cleaved by signal peptidase, which is active only on the luminal side of the ER membrane.
Use these criteria to decide whether a protein is translocated across
rough microsomal membranes. he mRNA is translated into protein in
a cell-free system in the absence or presence of microsomes. Samples
of newly synthesized proteins are treated in four diferent ways: (1) no
treatment, (2) addition of a protease, (3) addition of a protease and detergent, and (4) disruption of microsomes and addition of endoglycosidase
H (endo H), which removes N-linked sugars that are added in the ER. An
electrophoretic analysis of these samples is shown in Figure 12–20.
A. Explain the experimental results that are seen in the absence of microsomes (Figure 12–20, lanes 1 to 4).
B. Using the three criteria outlined in the problem, decide whether the
experimental results in the presence of microsomes (Figure 12–20,
lanes 5 to 8) indicate that the protein is translocated across microsomal
Figure 12–19 Arrangement of a
multipass transmembrane protein in the
ER membrane (Problem 12–102). Blue
hexagons represent covalently attached
oligosaccharides. The positions of
positively and negatively charged amino
acids lanking the second transmembrane
segment are shown.
THE ENDOPLASMIC RETICULUM
255
MICROSOMES
PRESENT
MICROSOMES
ABSENT
TREATMENT
protease
–
+
+
–
–
+
+
detergent
–
–
+
–
–
–
+
–
endo H
–
–
–
+
–
–
–
+
1
2
3
4
5
6
7
8
–
membranes. How would you account for the migration of the proteins in
Figure 12–20, lanes 5, 6, and 8?
C. Is the protein anchored in the membrane, or is it translocated all the way
through the membrane?
12–107 hings are not going well. You’ve just had a brief but tense meeting with
your research advisor that you will never forget, and it is clear that your
future in his lab is in doubt. He is not fond of your habit of working late
and sleeping late; he thinks it is at the root of your lack of productivity
(as he perceives it). A few days after the meeting, you roll in at the bright
and early hour of noon, to be informed by your student colleagues that
your advisor is “looking all over for you.” You are certain the axe is going
to fall. But it turns out he is excited, not upset. He’s just heard a seminar
that reminded him of the note you’d left on his desk the month before,
describing a selection scheme you had crafted (late one night, of course)
for isolating mutants in the ER translocation machinery. You are completely labbergasted.
You quickly settle on the details for the selection, which involves fusing an ER import signal to the N-terminus of the His4 gene product. His4
is a cytosolic enzyme that converts histidinol to histidine. Yeast strains
that are defective for an early step in the histidine biosynthetic pathway
can grow on added histidinol, if His4 is present. You decide to look for
temperature-sensitive (ts) mutants, which are normal at 24°C but defective at 37°C. Using a strain that expresses His4 with an ER import signal,
you select for cells that grow on histidinol at 30°C and screen them for
ones that die at 37°C. he irst mutant you isolate is in the Sec61 gene
(later shown to encode a principal component of the translocator through
which ribosomes insert nascent proteins across the ER membrane.) You
are back in your advisor’s good graces!
A. Why is it that normal cells with the modiied His4 cannot grow on histidinol, whereas cells with a defective ER-import apparatus can?
B. Why did the selection scheme set out to ind ts mutants? Why was selection applied at the intermediate temperature of 30°C, rather than at 24°C
or 37°C?
12–108 A classic paper describes a genetic method for determining the organi-
zation of a bacterial protein in the membrane of E. coli. he hydropathy
plot of the protein in Figure 12–21 indicated three potential membranespanning segments. Hybrid fusion proteins of diferent lengths, some
with internal deletions, were made with the membrane protein at the
N-terminus and alkaline phosphatase at the C-terminus (Figure 12–22).
Alkaline phosphatase is easy to assay in whole cells and has no signiicant
Figure 12–20 Results of translation of a
pure mRNA in the presence and absence
of microsomal membranes (Problem
12–106). Treatments of the products
of translation before electrophoresis
are indicated at the top of each lane.
Electrophoresis was on an SDS
polyacrylamide gel, which separates
proteins on the basis of size, with lower
molecular weight proteins migrating
farther down the gel.
Chapter 12: Intracellular Compartments and Protein Sorting
hydropathy index
256
Figure 12–21 hydropathy plot of a
membrane protein (Problem 12–108). The
three hydrophobic peaks indicate the
positions of three potential membranespanning segments.
more hydrophobic
0
more hydrophilic
0
50
100
150
200
amino acid number
250
300
hydrophobic stretches. Moreover, when it is on the cytoplasmic side of
the membrane its activity is low, and when it is on the external side of
the membrane (in the periplasmic space) its activity is high. he assayed
levels of alkaline phosphatase activity are indicated (HIGH or LOW) in
Figure 12–22.
A. How is the protein organized in the membrane? Explain how the results
with the fusion proteins indicate this arrangement.
B. How is the organization of the membrane protein altered by the deletion?
Are your measurements of alkaline phosphatase activity in the internally
deleted plasmids consistent with the altered arrangement?
12–109 Mitochondria and peroxisomes, as opposed to most other cellular mem-
branes, acquire new phospholipids via phospholipid exchange proteins.
One such protein, PC exchange protein, speciically transfers phosphatidylcholine (PC) between membranes. Its activity can be measured by
mixing red blood cell ghosts (intact plasma membranes with cytoplasm
removed) with synthetic phospholipid vesicles containing radioactively
labeled PC in both monolayers of the vesicle bilayer. After incubation
at 37°C, the mixture is centrifuged briely so that ghosts form a pellet,
whereas the vesicles stay in the supernatant. he amount of exchange is
determined by measuring the radioactivity in the pellet.
Figure 12–23 shows the result of an experiment along these lines,
using labeled (donor) vesicles with an outer radius of 10.5 nm and a
bilayer 4.0 nm in thickness. No transfer occurred in the absence of the
exchange protein, but in its presence up to 70% of the labeled PC in the
vesicles could be transferred to the red cell membranes.
Several control experiments were performed to explore the reason
why only 70% of the label in donor vesicles was transferred.
original
constructs
1
alkaline phosphatase activity
34
HIGH
55
HIGH
2
105
3
LOW
131
4
LOW
225
HIGH
5
310
6
internally deleted
derivatives
105
HIGH
3*
4*
HIGH
131
HIGH
225
LOW
5*
310
6*
deletion of
residues 68–103
LOW
Figure 12–22 Structures of hybrid
proteins used to determine the
organization of a membrane protein
(Problem 12–108). The membrane protein
(orange segment) is at the N-terminus
and alkaline phosphatase (blue segment)
is at the C-terminus of the protein. The
inverted V indicates the site from which
amino acids were deleted from modiied
hybrid proteins. The most C-terminal
amino acid of the membrane protein is
numbered in each hybrid protein. The
activity of alkaline phosphatase in each
hybrid protein is shown on the right.
THE ENDOPLASMIC RETICULUM
MCAT STYLE
Passage 1 (Questions 12–110 to 12–112)
You are studying a transcription factor that controls entry into the cell cycle. It is
present in the nucleus in nondividing cells, but rapidly exits the nucleus when
cells are stimulated to begin cell division. You hypothesize that the transcription
factor represses expression of genes that drive entry into the cell cycle, and that
transport of the repressor out of the nucleus is a critical step that helps initiate the
cell cycle. While searching for the signals that control this step, you ind that the
protein is phosphorylated by a protein kinase at a speciic serine. Since phosphorylation of proteins is an important mode of regulation, you investigate further. To
do so, you create a mutant version of the transcription factor in which the serine
that is the target for phosphorylation is changed to an alanine, which cannot be
phosphorylated. When you express this mutant version of the transcription factor
in the cell, you ind that very little of it can be detected in the nucleus.
12–110 Transport of the transcription factor into the nucleus most likely requires:
I. Binding to a nuclear import receptor
II. Direct participation of Ran-GDP
III. Involvement of the Sec61 complex
A. I
B. I and II
C. I and III
D. I, II, and III
12–111 Which one of the following hypotheses best explains how phosphoryla-
tion regulates nuclear localization of the transcription factor?
A. Phosphorylation causes a conformational change that exposes a nuclear
export signal.
B. Phosphorylation near a nuclear export signal blocks binding of a nuclear
export factor.
C. Phosphorylation near a nuclear localization signal blocks binding of a
nuclear import factor.
D. Phosphorylation promotes binding of the transcription factor to RanGTP.
12–112 How might you test the best hypothesis from the previous question?
A. Addition of a nuclear import signal to the mutant transcription factor
should allow it to accumulate in the nucleus.
B. Deletion of the nuclear export signal from the mutant transcription factor should allow it to accumulate in the nucleus.
100
labeled PC removed
from vesicles (percent)
1. Five times as many membranes from red cell ghosts were included in the
incubation: the transfer still stopped at the same point.
2. Fresh exchange protein was added after 1 hour: it caused no further
transfer.
3. he labeled lipids remaining in donor vesicles at the end of the reaction
were extracted and made into fresh vesicles: 70% of the label in these
vesicles was exchangeable.
When the red cell ghosts that were labeled in this experiment were
used as donor membranes in the reverse experiment (that is, transfer of
PC from red cell membranes to synthetic vesicles), 96% of the label could
be transferred to the acceptor vesicles.
A. What possible explanations for the 70% limit do each of the three control
experiments eliminate?
B. What do you think is the explanation for the 70% limit? (Hint: the area of
the outer surface of these small donor vesicles is about 2.6 times larger
than the area of the inner surface.)
C. Why do you think that almost 100% of the label in the red cell membrane
can be transferred back to the vesicle?
257
+ PC exchange protein
80
60
40
20
0
– PC exchange protein
0
0.5
1.0
time (hours)
1.5
Figure 12–23 Transfer of labeled PC from
donor vesicles to red cell membranes by
PC exchange protein (Problem 12–109).
258
Chapter 12: Intracellular Compartments and Protein Sorting
C. Deletion of the Ran-GDP binding site from the mutant transcription factor should allow it to accumulate in the nucleus.
D. Inactivation of the Ran-GTPase should allow the normal—but not the
mutant—transcription factor to enter the nucleus.
Passage 2 (Questions 12–113 to 12–115)
When cells are broken open, the endoplasmic reticulum (ER) fragments into
spherical vesicles coated with ribosomes, which are called rough microsomes.
hese rough microsomes can be puriied and used to form a lipid bilayer that
separates two liquid-illed chambers. One side of the lipid bilayer is coated with
ribosomes and corresponds to the cytosolic side of the ER, while the other corresponds to the luminal side of the ER. he chambers on each side of the membrane can be monitored to detect conductance of salt ions across the membrane.
No conductance can be detected across the rough microsomal membrane as it is
isolated from cells. However, treatment of the ribosome-coated side of the membrane with puromycin, a compound that causes release of growing peptide chains
from ribosomes, causes a large increase in conductance. Further treatment with
high-salt bufers causes the membrane to be impermeable to ions again.
12–113 Association of ribosomes with the endoplasmic reticulum requires:
I. Signal-recognition particle
II. Chaperones
III. Protein translation
A. I
B. I and III
C. II and III
D. I, II, and III
12–114 Which one of the following statements best explains the increase in con-
ductance caused by puromycin?
A. Premature termination of peptide synthesis by puromycin leads to
improperly folded proteins that activate the unfolded protein response.
B. Release of the growing peptide exposes the water-illed pore that is used
for translocation of proteins across the ER membrane.
C. Release of the peptide allows the signal-recognition particle to open a
channel for translocation of peptides across the ER membrane.
D. Treatment with puromycin opens channels in the membrane that are
normally used to release Ca2+ from the ER.
12–115 Why does washing the membrane with a bufer containing high concen-
trations of salt block the conductance of the microsomal membrane?
A. High salt causes the Ca2+ channel to close.
B. High salt increases the resistance of the membrane.
C. High-salt removal of ribosomes closes the pore.
D. High salt shuts of the unfolded protein response.
Chapter 13
259
CHAPTER
Intracellular Membrane Traffic
THE MOLECULAR MECHANISMS OF MEMBRANE
TRANSPORT AND THE MAINTENANCE OF
COMPARTMENTAL DIVERSITY
TERMS TO LEARN
adaptor protein
COPI-coated vesicle
ARF protein
COPII-coated vesicle
cargo
dynamin
clathrin
lumen
clathrin-coated vesicle NSF
coated vesicle
phosphatidylinositide
coat-recruitment
(PIP)
GTPase
Rab cascade
Rab effector
Rab protein
Sar1 protein
SNARE protein (SNARE)
transport vesicle
t-SNARE
v-SNARE
DEFINITIONS
Match each deinition below with its term from the list above.
13–1
General term for a membrane-enclosed container that moves material
between membrane-enclosed compartments within the cell.
13–2
Any of a large family of monomeric GTPases present in the plasma membrane and organelle membranes that confer speciicity on vesicle docking.
13–3
A protein that mediates binding between the clathrin coat and transmembrane proteins, including transmembrane cargo receptors.
13–4
Cytosolic GTPase that binds to the neck of a clathrin-coated vesicle and
helps it to pinch of from the membrane.
13–5
he protein that catalyzes the disassembly of the helical domains of
paired SNARE proteins.
13–6
Coated vesicle that transports material from the plasma membrane and
between endosomal and Golgi compartments.
13–7
he coat-recruitment GTPase responsible for both COPI coat assembly
and clathrin coat assembly at Golgi membranes.
13–8
Protein that facilitates vesicle transport, docking, and membrane fusion
once it is bound by an activated Rab protein.
13–9
General term for a member of the large family of proteins that catalyze
the membrane fusion reactions in membrane transport.
13–10
he interior space of a membrane-enclosed compartment.
13–11
General term for a transport vesicle that carries a distinctive cage of
proteins covering its cytosolic surface.
13
IN THIS CHAPTER
THE MOLECULAR MECHANISMS
OF MEMBRANE TRANSPORT
AND THE MAINTENANCE OF
COMPARTMENTAL DIVERSITY
TRANSPORT FROM THE ER
THROUGH THE
GOLGI APPARATUS
TRANSPORT FROM THE
TRANS GOLGI NETWORK TO
LYSOSOMES
TRANSPORT INTO THE
CELL FROM THE PLASMA
MEMBRANE: ENDOCYTOSIS
TRANSPORT FROM THE TRANS
GOLGI NETWORK TO THE CELL
EXTERIOR: EXOCYTOSIS
260
Chapter 13: Intracellular Membrane Traffic
13–12
he coat-recruitment GTPase responsible for COPII coat assembly at the
ER membrane.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
13–13
In all events involving fusion of a vesicle to a target membrane, the cytosolic lealets of the vesicle and target bilayers always fuse together, as do
the lealets that are not in contact with the cytosol.
13–14
Complementary Rab proteins on transport vesicles and target membranes bind to one another to allow transport vesicles to dock selectively
at their appropriate target membranes.
THOUGHT PROBLEMS
13–15
In a nondividing cell such as a liver cell, why must the low of membrane
between compartments be balanced, so that the retrieval pathways
match the outward low? Would you expect the same balanced low in a
gut epithelial cell, which is actively dividing?
he diagram in Figure 13–1 shows the various intracellular compartments involved in the biosynthetic-secretory, endocytic, and retrieval
pathways.
A. Label the various compartments in the diagram.
B. Indicate on the arrows whether the indicated low is part of the biosynthetic-secretory pathway, the endocytic pathway, or a retrieval pathway.
13–16
13–17
Discuss the following analogy: “Cargo receptors competing to be transported by the coated-pit system can be compared to skiers joining a
cable-car network. Entry is permitted to ticket holders only, but there
is no guarantee of who is found with whom in a particular cable car,
although all travelers, hopefully, will reach the next station.”
13–18
he clathrin coat on a vesicle is made up of numerous triskelions that
form a cage 60–200 nm in diameter, composed of both pentagonal and
hexagonal faces, just like C60 fullerene (Figure 13–2). Sketch the location
of an individual triskelion in the clathrin-coated vesicle in Figure 13–2B.
CYTOSOL
EXTRACELLULAR SPACE
Figure 13–1 The intracellular
compartments in the biosyntheticsecretory, endocytic, and retrieval
pathways, with low between
compartments indicated (Problem 13–16).
THE MOLECULAR MECHANISMS OF MEMBRANE TRANSPORT
(A)
(B)
261
Figure 13–2 Structure of a clathrin coat
(Problem 13–18). (A) A triskelion subunit.
(B) A clathrin-coated vesicle. (C) A C60
fullerene.
(C)
25 nm
triskelion
clathrin-coated vesicle
fullerene (C60)
At what point in its structure does a triskelion have to be most lexible to
accommodate changes in size of the vesicle? At what point does it have to
be most lexible to it into both the pentagonal and hexagonal faces?
13–19
Yeast, and many other organisms, make a single type of clathrin heavy
chain and a single type of clathrin light chain; thus, they make a single
kind of clathrin coat. How is it, then, that a single clathrin coat can be
used for three diferent transport pathways—Golgi to late endosomes,
plasma membrane to early endosomes, and immature secretory vesicles
to Golgi—that each involves diferent specialized cargo proteins?
13–20
Imagine that ARF1 protein was mutated so that it could not hydrolyze
GTP, regardless of its binding partners. Would you expect COPI-coated
vesicles to form normally? How would you expect transport mediated by
COPI-coated vesicles to be afected? If this were the only form of ARF1 in
a cell, would you expect it to be lethal? Explain your answers.
13–21
How can it possibly be true that complementary pairs of speciic SNAREs
uniquely mark vesicles and their target membranes? After vesicle fusion,
the target membrane will contain a mixture of t-SNAREs and v-SNAREs.
Initially, these SNAREs will be tightly bound to one another, but NSF can
pry them apart, reactivating them. What do you suppose prevents target
membranes from accumulating a population of v-SNAREs equal to or
greater than their population of t-SNAREs?
13–22
Viruses are the ultimate scavengers—a necessary consequence of their
small genomes. Wherever possible, they make use of the cell’s machinery to accomplish the steps involved in their own reproduction. Many
diferent viruses have membrane coverings. hese so-called enveloped
viruses gain access to the cytosol by fusing with a cell membrane. Why
do you suppose that each of these viruses encodes its own special fusion
protein, rather than making use of a cell’s SNAREs?
CALCULATIONS
13–23
For fusion of a vesicle with its target membrane to occur, the membranes
have to be brought to within 1.5 nm so that the two bilayers can join
(Figure 13–3). Assuming that the relevant portions of the two membranes at the fusion site are circular regions 1.5 nm in diameter, calculate
the number of water molecules that would remain between the membranes. (Water is 55.5 M and the volume of a cylinder is πr2h.) Given that
an average phospholipid occupies a membrane surface area of 0.2 nm2,
how many phospholipids would be present in each of the opposing monolayers at the fusion site? Are there suicient water molecules to bind
to the hydrophilic head groups of this number of phospholipids? (It is
estimated that 10–12 water molecules are normally associated with each
phospholipid head group at the exposed surface of a membrane.)
vesicle
target membrane
1.5 nm
Figure13–3 Close approach of a vesicle
and its target membrane in preparation
for fusion (Problem 13–23).
262
Chapter 13: Intracellular Membrane Traffic
GTP
COPI
ARF-GDP
(cytosol)
ARF-GTP
(membrane)
COPI-coated
vesicles
GDP
GEF-catalyzed
DATA HANDLING
13–24
When the fungal metabolite brefeldin A is added to cells, the Golgi apparatus largely disappears and the Golgi proteins intermix with those in
the ER. Brefeldin-A treatment also causes the rapid dissociation of some
Golgi-associated peripheral membrane proteins, including subunits of
the COPI coat. hese observations imply that brefeldin A prevents transport involving COPI-coated vesicles by blocking the assembly of coats,
and thus the budding of transport vesicles. In principle, brefeldin A
could block formation of COPI-coated vesicles at any point in the normal scheme for assembly, which is shown in Figure 13–4. he following
observations identify the point of action of brefeldin A.
1. ARF with bound GTPγS (a nonhydrolyzable analog of GTP) causes
COPI-coated vesicles to form when added to Golgi membranes. Formation of vesicles in this way is not afected by brefeldin A.
2. ARF with bound GDP exchanges the GDP for GTP when added to
Golgi membranes. his exchange reaction does not occur in the presence
of brefeldin A. he essential component in the Golgi membrane is sensitive to trypsin digestion, suggesting that it is a protein.
Given these experimental observations, how do you think brefeldin A
blocks formation of COPI-coated vesicles?
13–25
Small GTPases are generally active in the GTP-bound state and inactive
when the GTP is hydrolyzed to GDP. In the absence of a GTPase-activating protein (GAP), small GTPases typically hydrolyze GTP very slowly.
he mechanism by which a GAP stimulates GTP hydrolysis is known for
the small GTPase Ras. When Ras-GAP binds to Ras, it alters the conformation of Ras and provides a critical, catalytic arginine “inger” that stabilizes the transition state for GTP hydrolysis, thereby stimulating hydrolysis by several orders of magnitude.
During assembly of COPI-coated vesicles, ARF1—a small GTPase—
binds to ARF1-GAP, which locks ARF1 into its active catalytic conformation but does not supply the catalytic arginine. Since COPI subunits also
bind to ARF1, you wonder if they might afect GTP hydrolysis. To test this
possibility, you mix ARF1, ARF1-GAP, and COPI subunits in various combinations and measure GTP hydrolysis (Table 13–1).
How would you interpret these results? How might you further test
your conclusions?
13–26
SNAREs exist as complementary partners that carry out membrane
fusions between appropriate vesicles and their target membranes. In this
way, a vesicle with a particular variety of v-SNARE will fuse only with a
TABLE 13–1 Rates of GTP hydrolysis by various combinations of
ARF1, ARF1-GAP, and COPI subunits (Problem 13–25).
Components added
Rate of GTP hydrolysis
ARF1
0
ARF1 + ARF1-GAP
1
ARF1 + COPI subunits
0
ARF1 + ARF1-GAP + COPI subunits
1000
Figure 13–4 Normal pathway for
formation of CoPI-coated vesicles
(Problem 13–24). The small GTPase ARF
carries a bound GDP in its cytosolic
form. In response to a guanine nucleotide
exchange factor (GEF), ARF releases GDP
and picks up GTP. Binding of GTP causes
a conformational change that exposes
a fatty acid tail on ARF, which promotes
binding of ARF-GTP to the membrane.
CoPI subunits bind to ARF-GTP to form
CoPI-coated vesicles.
THE MOLECULAR MECHANISMS OF MEMBRANE TRANSPORT
stra i n A
stra i n B
v
t
t
v
pro-Pase
(B)
100
protease
alkaline phosphatase
(% maximum)
(A)
263
DOCKING
pro-Pase
protease
FUSION
Pase
SNARE
combinations
75
50
25
0
strain A vt
strain B vt
experiment 1
t
t
2
t
vt
3
membrane that carries the complementary t-SNARE. In some instances,
however, fusions of identical membranes (homotypic fusions) are known
to occur. For example, when a yeast cell forms a bud, vesicles derived
from the mother cell’s vacuole move into the bud where they fuse with
one another to form a new vacuole. hese vesicles carry both v-SNAREs
and t-SNAREs. Are both types of SNAREs essential for this homotypic
fusion event?
To test this point, you have developed an ingenious assay for fusion
of vacuolar vesicles. You prepare vesicles from two diferent mutant
strains of yeast: strain B has a defective gene for vacuolar alkaline phosphatase (Pase); strain A is defective for the protease that converts the
precursor of alkaline phosphatase (pro-Pase) into its active form (Pase)
(Figure 13–5A). Neither strain has active alkaline phosphatase, but when
extracts of the strains are mixed, vesicle fusion generates active alkaline
phosphatase, which can be easily measured (Figure 13–5).
Now you delete the genes for the vacuolar v-SNARE, t-SNARE, or both
in each of the two yeast strains. You prepare vacuolar vesicles from each
and test them for their ability to fuse, as measured by the alkaline phosphatase assay (Figure 13–5B).
What do these data say about the requirements for v-SNAREs and
t-SNAREs in the fusion of vacuolar vesicles? Does it matter which kind of
SNARE is on which vesicle?
13–27
You wish to identify the target proteins that are bound by NSF and its two
accessory proteins. You incubate puriied NSF and its accessory proteins
with a crude detergent extract of synaptic membranes, and then add
NSF-speciic antibodies that are attached to beads. By centrifuging the
mixture, you can readily separate the beads, and any attached proteins,
from the rest of the crude extract. he proteins attached to the beads can
be analyzed by SDS polyacrylamide-gel electrophoresis. When the incubation is carried out in the presence or absence of ATP, you ind that NSF
alone is present on the beads. If you incubate in the presence of ATPγS, a
nonhydrolyzable analog of ATP, the beads bring down NSF, its accessory
proteins, syntaxin, SNAP25, and synaptobrevin. What is the substrate for
NSF and its accessory proteins? Why does the experiment work when
ATPγS is present, but not in the presence or absence of ATP?
13–28
he Sec4 gene of budding yeast encodes a small GTPase that plays
an essential role in the secretion pathway that forms the daughter
bud. Normally, about 80% of the Sec4 protein is found on the cytosolic surface of transport vesicles and 20% is free in the cytosol. When
vt
t
4
v
v
5
v
vt
6
vt
v
7
t
v
8
v
t
9
–
vt
10
vt
–
11
Figure 13–5 SNARE requirements for
vesicle fusion (Problem 13–26).
(A) Scheme for measuring the fusion of
vacuolar vesicles. (B) Results of fusions
of vesicles with different combinations of
v-SNAREs and t-SNAREs. The SNAREs
present on the vesicles of the two strains
are indicated as v (v-SNARE) and
t (t-SNARE).
Chapter 13: Intracellular Membrane Traffic
264
temperature-sensitive Sec4 mutants of yeast (Sec4ts) are incubated at
high temperature, growth ceases and small vesicles accumulate in the
daughter bud.
To deine the role of Sec4 in secretion, you engineer two speciic Sec4
mutants based on the way other small GTPases work. One mutant, Sec4ccΔ, lacks two cysteines at its C-terminus, which you expect will prevent
attachment of the fatty acid required for membrane binding. he second
mutant, Sec4N133I, encodes an isoleucine in place of the normal asparagine at position 133; you expect that this protein will be locked into its
active state, even though it should not be able to bind GTP or GDP.
You ind that Sec4-ccΔ binds GTP but remains entirely cytosolic with
none bound to vesicles. When expressed at high levels in yeast that also
have a normal Sec4 gene, it does not inhibit their growth. In contrast,
Sec4N133I is located almost entirely on vesicles, and when it is expressed
at high levels in normal yeast, it completely inhibits growth, and the yeast
are found to be packed with small vesicles.
A. Do you think Sec4 is required for formation of vesicles, for vesicle fusion
with target membranes, or for both? Based on its function, would you
guess it was analogous to mammalian ARF, Sar1, or Rab proteins?
B. Using your knowledge of the way the analogous mammalian protein
works, outline how you think normal Sec4 functions in vesicle formation
and fusion. Why is some Sec4 free in the cytosol of wild-type cells? How
does removal of the C-terminal cysteines prevent Sec4-ccΔ from carrying out its function?
C. Why do you think expression of Sec4N133I inhibits growth of the yeast
that also express normal Sec4?
TRANSPORT FROM THE ER THROUGH THE GOLGI
APPARATUS
TERMS TO LEARN
cis face
cis Golgi network (CGN)
cisternal maturation model
complex oligosaccharide
Golgi apparatus (Golgi complex)
high-mannose oligosaccharide
O-linked glycosylation
proteoglycan
trans face
trans Golgi network (TGN)
vesicle transport model
DEFINITIONS
Match each deinition below with its term from the list above.
13–29
he hypothesis that new cisternae form continuously at the cis face of the
Golgi and then migrate through the stack as they mature.
13–30
Molecule consisting of one or more glycosaminoglycan chains attached
to a core protein.
13–31
he side of the Golgi stack at which material enters the organelle.
13–32
Chain of sugars attached to a glycoprotein that is generated by initially
trimming the original oligosaccharide attached in the ER and by then
adding other sugars.
13–33
Membrane-enclosed organelle in eukaryotic cells in which proteins and
lipids transferred from the ER are modiied and sorted.
13–34
Chain of sugars attached to a glycoprotein that contains many mannose
residues.
13–35
Meshwork of interconnected cisternae and tubules on the side of the
Golgi stack at which material is transferred out of the Golgi.
TRANSPORT FROM THE ER THROUGH THE GOLGI APPARATUS
265
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
13–36
here is one strict requirement for the exit of a protein from the ER: it
must be correctly folded.
13–37
All of the glycoproteins and glycolipids in intracellular membranes have
their oligosaccharide chains facing the lumenal side, and all those in the
plasma membrane have their oligosaccharide chains facing the outside
of the cell.
13–38
he Golgi apparatus confers the heaviest glycosylation of all on proteoglycan core proteins, which are converted into proteoglycans by the
addition of one or more O-linked glycosaminoglycan chains.
THOUGHT PROBLEMS
13–39
How is it that soluble proteins in the ER can be selectively recruited into
vesicles destined for the Golgi?
13–40
Isn’t quality control always a good thing? How can quality control in the
ER be detrimental to cystic ibrosis patients?
13–41
he C-terminal 40 amino acids of three ER resident proteins—calnexin,
calreticulin, and HMG CoA reductase—are shown in Figure 13–6. Decide
for each protein whether it is likely to be a transmembrane protein or a
soluble one. Explain your answers.
13–42
If you were to remove the ER retrieval signal from protein disulide
isomerase (PDI), which is normally a soluble resident of the ER lumen,
where would you expect the modiied PDI to be located?
13–43
he KDEL receptor must shuttle back and forth between the ER and the
Golgi apparatus in order to accomplish its task of ensuring that soluble
ER proteins are retained in the ER lumen. In which compartment does
the KDEL receptor bind its ligands more tightly? In which compartment
does it bind its ligands more weakly? What is thought to be the basis
for its diferent binding ainities in the two compartments? If you were
designing the system, in which compartment would you have the highest concentration of KDEL receptor? Would you predict that the KDEL
receptor, which is a transmembrane protein, would itself possess an ER
retrieval signal?
13–44
When the KDEL retrieval signal is added to rat growth hormone or
human chorionic gonadotropin, two proteins that are normally secreted,
the proteins are still secreted, but about six times more slowly. If the
C-terminal L in the signal is changed to V, the proteins are once again
secreted at their normal rate. By contrast, bona ide ER resident proteins
rarely, if ever, are secreted from the cell; they are usually captured and
returned very eiciently. How is it, do you suppose, that normal resident
proteins with a KDEL signal are eiciently retained in the ER, whereas
secreted proteins to which a KDEL signal has been added are not
eiciently retained? Is this what you would expect if the KDEL signal and
the KDEL receptor accounted entirely for retention of soluble proteins in
the ER?
Calnexin
C-terminus
...KDKGDEEEEGEEKLEEKQKSDAEEDGGTVSQEEEDRKPKAEEDEILNRSPRNRKPRRE
Calreticulin
...KQDEEQRLKEEEEDKKRKEEEEAEDKEDDEDKDEDEEDEEDKEEDEEEDVPGQAKDEL
HMG CoA reductase
...PGENARQLARIVCGTVMAGELSLMAALAAGHLVKSHMIHNRSKINLQDLQGACTKKTA
Figure 13–6 C-terminal amino acids of
proteins that are residents of the ER
(Problem 13–41).
Chapter 13: Intracellular Membrane Traffic
266
membrane-spanning
segment
13–45
pMS20
K S S I A S F F F I I G L I I G L F L V L R
pMS18
K S S I A S F F F I I G L - - G L F L V L R
pMS16
K S S I A S F F F I I G - - - - L F L V L R
pMS14
K S S I A S F F F I I - - - - - - F L V L R
pMS12
K S S I A S F F F I - - - - - - - - L V L R
pMS8
K S S I A - - - - - - - - - - - - F L V L R
pMS0
K - - - - - - - - - - - - - - - - - - - - R
Cells have evolved a set of complicated pathways for addition of carbohydrates to proteins, implying that carbohydrates serve important functions. List at least three functions that carbohydrates on proteins are
known to carry out.
DATA HANDLING
he vesicular stomatitis virus (VSV) G protein is a typical membrane glycoprotein. In addition to its signal peptide, which is removed after import
into the ER, the G protein contains a single membrane-spanning segment
that anchors the protein in the plasma membrane. he membrane-spanning segment consists of 20 uncharged and mostly hydrophobic amino
acids that are lanked by basic amino acids (Figure 13–7). Twenty amino
acids arranged in an α helix are just suicient to span the 3-nm thickness
of the lipid bilayer of the membrane.
To test the length requirements for membrane-spanning segments,
you modify a cloned version of the VSV G protein to generate a series of
mutants in which the membrane-spanning segment is shorter, as indicated in Figure 13–7. When you introduce the modiied plasmids into
cultured cells, roughly the same amount of G protein is synthesized from
each mutant as from wild-type cells. You analyze the cellular distribution
of the altered G proteins in several ways.
1. You examine the cellular location of the modiied VSV G proteins by
immunoluorescence microscopy, using G-protein-speciic antibodies
tagged with luorescein.
2. You characterize the attached oligosaccharide chains by digestion
with endoglycosidase H (Endo H), which cleaves of N-linked oligosaccharides until the irst mannose is removed in the medial portion of the
Golgi apparatus (see Figure 13–8, Problem 13–47).
3. You determine whether the altered VSV G proteins retain the small
C-terminal cytoplasmic domain (which characterizes the normal G protein) by treating isolated microsomes with a protease. In the normal VSV
G protein, this domain is sensitive to protease treatment and is removed.
he results of these experiments are summarized in Table 13–2.
A. To the extent these data allow, deduce the intracellular location of each
altered VSV G protein that fails to reach the plasma membrane.
B. For the VSV G protein, what is the minimum length of the membranespanning segment that is suicient to anchor the protein in the membrane?
C. What is the minimum length of the membrane-spanning segment that is
consistent with proper sorting of the G protein? How is it that transmembrane segments shorter than this can make it to the Golgi apparatus, but
then not be able to exit?
13–46
13–47
You have isolated several mutant cell lines that are defective in their ability to add carbohydrate to exported proteins. Using an easily puriied
Figure 13–7 The membrane-spanning
domains of normal and mutant VSV
G proteins (Problem 13–46). Plasmid
numbers indicate the number of amino
acids in the membrane-spanning
segment; for example, pMS20 contains
the wild type, 20-amino-acid segment.
Dashed lines indicate amino acids
that are missing in the other plasmids.
Highlighted letters indicate the basic
amino acids that lank the membranespanning segment.
TRANSPORT FROM THE ER THROUGH THE GOLGI APPARATUS
267
TABLE 13–2 Results of experiments characterizing the cellular distribution of
G proteins from normal and mutant cells (Problem 13–46).
Plasmid
Cellular location
Endo H treatment
Protease treatment
pMS20
plasma membrane
resistant
sensitive
pMS18
plasma membrane
resistant
sensitive
pMS16
plasma membrane
resistant
sensitive
pMS14
plasma membrane
resistant
sensitive
pMS12
intracellular
+/– resistant
sensitive
pMS8
intracellular
sensitive
sensitive
pMS0
intracellular
sensitive
resistant
protein that carries only N-linked complex oligosaccharides, you have
analyzed the sugar monomers that are added in the diferent mutant
cells. Each mutant is unique in the kinds and numbers of diferent sugars
contained in its N-linked oligosaccharides (Table 13–3).
A. Arrange the mutants in the order that corresponds to the steps in the pathway for processing N-linked oligosaccharides (Figure 13–8). (Assume
that each mutant cell line is defective for a single enzyme required to
construct the N-linked oligosaccharide.)
B. Which of these mutants are defective in processing events that occur in
the ER? Which mutants are defective in processing events that occur in
the Golgi?
C. Which of the mutants are likely to be defective in a processing enzyme
that is directly responsible for modifying N-linked oligosaccharides?
Which mutants might not be defective in a processing enzyme, but rather
in another enzyme that afects oligosaccharide processing indirectly?
TABLE 13–3 Analysis of the sugars present in the N-linked oligosaccharides
from wild-type and mutant cell lines defective in oligosaccharide processing
(Problem 13–47).
Cell line
Man
GlcNAc
Gal
NANA
Glc
Wild type
3
5
3
3
0
Mutant A
3
5
0
0
0
Mutant B
5
3
0
0
0
Mutant C
9
2
0
0
3
Mutant D
9
2
0
0
0
Mutant E
5
2
0
0
0
Mutant F
3
3
0
0
0
Mutant G
8
2
0
0
0
Mutant H
9
2
0
0
2
Mutant I
3
5
3
0
0
Abbreviations: Man = mannose; GlcNAc = N-acetylglucosamine; Gal = galactose;
NANA = N-acetylneuraminic acid, or sialic acid; Glc = glucose. Numbers indicate the number
of sugar monomers in the oligosaccharide.
268
Chapter 13: Intracellular Membrane Traffic
glucosidase I
Golgi
mannosidase I
glucosidase II
GlcNAc transferase I
ER mannosidase
Asn
Asn
1
3
high-mannose
oligosaccharide
(2) GlcNAc transferase II
UDP
(3) galactose transferase
CMP
(3) NANA transferase
Asn
4
UDP + 3 CMP
Asn
5
Endo Hsensitive
Endo Hresistant
complex
oligosaccharide
GOLGI LUMEN
ER LUMEN
13–48
UDP
5
Asn
2
KEY:
= N-acetylglucosamine (GlcNAc)
UDP
UDP
Asn
Golgi
mannosidase II
= mannose (Man)
= glucose (Glc)
= galactose (Gal)
Two extreme models—vesicle transport and cisternal maturation—have
been proposed to account for the movement of molecules across the
polarized structure of the Golgi apparatus. In the vesicle transport model,
the individual Golgi cisternae remain in place as proteins move through
them (Figure 13–9A). By contrast, in the cisternal maturation model, the
individual Golgi cisternae move across the stack, carrying the proteins
with them (Figure 13–9B). Transport vesicles serve critical functions in
both models, but their roles are distinctly diferent. Describe the roles of
the transport vesicles in each of the two models. Comment speciically
on the roles of vesicles in the forward movement of proteins across the
Golgi stack, in the retention of Golgi resident proteins in individual cisternae, and on the return of escaped ER proteins to the ER.
= N-acetylneuraminic acid (sialic acid, or NANA)
Figure 13–8 oligosaccharide processing
in the ER and the Golgi apparatus
(Problem 13–47).
(A) VESICLE TRANSPORT MODEL
(B) CISTERNAL MATURATION MODEL
ER
vesicular
tubular cluster CGN
cisternae
cis
medial trans
TGN
Figure 13–9 Two models for the
movement of molecules through the
Golgi apparatus (Problem 13–48).
(A) The vesicle transport model. (B) The
cisternal maturation model. In (B), the
individual cisternae have been separated
for illustration purposes. ER, endoplasmic
reticulum; CGN, cis Golgi network;
TGN, trans Golgi network.
TRANSPORT FROM THE ER THROUGH THE GOLGI APPARATUS
269
TABLE 13–4 Addition of galactose to VSV G protein after fusion of
VSV-infected cells with uninfected cells (Problem 13–49).
Infected cells
Uninfected cells
Precipitate
Supernatant
Mutant cells
Wild-type cells
45%
55%
Mutant cells
Mutant cells
5%
95%
Wild-type cells
Wild-type cells
85%
15%
The percentage of radioactivity in the precipitate indicates the fraction of
GlcNac-labeled G protein that has acquired galactose.
One early test of the vesicle transport and cisternal maturation models
(see Figure 13–9) looked for the movement of a protein between Golgi
cisternae. his study made use of mutant cells that cannot add galactose
to proteins, which normally occurs in the trans compartment of the Golgi
(see Figure 13–8). he mutant cells were infected with vesicular stomatitis
virus (VSV) to provide a convenient marker protein, the viral G protein.
At an appropriate point in the infection, an inhibitor of protein synthesis
was added to stop further synthesis of G protein. he infected cells were
then incubated briely with a radioactive precursor of GlcNAc, which is
added only in the medial cisterna of the Golgi (see Figure 13–8). Next,
the infected mutant cells were fused with uninfected wild-type cells to
form a common cytoplasm containing both wild-type and mutant Golgi
stacks. After a few minutes, the cells were dissolved with detergent and
all the VSV G protein was captured using G-protein-speciic antibodies.
After separation from the antibodies, the G proteins carrying galactose
were precipitated, using a lectin that binds galactose. he radioactivity in
the precipitate and in the supernatant was measured. he results of this
experiment along with control experiments (which used mutant cells
only or wild-type cells only) are shown in Table 13–4.
A. Between which two compartments of the Golgi apparatus is the movement of proteins being tested in this experiment? Explain your answer.
B. If proteins moved through the Golgi apparatus by cisternal maturation,
what would you predict for the results of this experiment? If proteins
moved through the Golgi via vesicular transport, what would you predict?
C. Which model is supported by the results in Table 13–4?
13–49
MEDICAL LINKS
13–50
Processing of N-linked oligosaccharides is not uniform among species.
Most mammals, with the exception of humans and Old World primates,
occasionally add galactose—in place of an N-acetylneuraminic acid—
to a galactose, forming a terminal Gal(α1–3)Gal disaccharide on some
branches of an N-linked oligosaccharide. How does this explain the preferred use of Old World primates for production of recombinant proteins
for therapeutic use in humans?
TRANSPORT FROM THE TRANS GOLGI NETWORK TO
LYSOSOMES
TERMS TO LEARN
acid hydrolase
lysosomal storage disease
autophagosome
lysosome
autophagy
M6P receptor protein
vacuole
Chapter 13: Intracellular Membrane Traffic
270
DEFINITIONS
Match each deinition below with its term from the list above.
13–51
Digestion of obsolete parts of the cell by the cell’s own lysosomes.
13–52
Very large, luid-illed vesicle found in most plant and fungal cells, typically occupying more than 30% of the cell volume.
13–53
Membrane-enclosed organelle in eukaryotic cells that contains digestive
enzymes, which are typically most active at the acidic pH found in the
lumen.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
13–54
Lysosomal membranes contain a proton pump that utilizes the energy of
ATP hydrolysis to pump protons out of the lysosome, thereby maintaining the lumen at a low pH.
13–55
Late endosomes are converted to mature lysosomes by the loss of distinct
endosomal membrane proteins and a further decrease in their internal
pH.
13–56
If cells were treated with a weak base such as ammonia or chloroquine,
which raises the pH of organelles toward neutrality, M6P receptors would
be expected to accumulate in the Golgi because they could not bind to
the lysosomal enzymes.
THOUGHT PROBLEMS
13–57
How does the low pH of lysosomes protect the rest of the cell from lysosomal enzymes in case the lysosome breaks?
13–58
Imagine that an autophagosome is formed by engulfment of a mitochondrion by the ER membrane. How many layers of membrane would
separate the matrix of the mitochondrion from the cytosol outside the
autophagosome? Identify the source of each membrane and the spaces
between the membranes.
he principal pathway for transport of lysosomal hydrolases from the
trans Golgi network (pH 6.6) to the late endosomes (pH 6) and for the
recycling of M6P receptors back to the Golgi depends on the pH diference between those two compartments. From what you know about M6P
receptor binding and recycling and the pathways for delivery of material
to lysosomes, describe the consequences of changing the pH in those
two compartments.
A. What do you suppose would happen if the pH in late endosomes were
raised to pH 6.6?
B. What do you suppose would happen if the pH in the trans Golgi network
were lowered to pH 6?
13–59
13–60
Melanosomes are specialized lysosomes that store pigments for eventual release by exocytosis. Various cells such as skin and hair cells then
take up the pigment, which accounts for their characteristic pigmentation. Mouse mutants that have defective melanosomes often have pale
or unusual coat colors. One such light-colored mouse, the Mocha mouse
(Figure 13–10), has a defect in the gene for one of the subunits of the
adaptor protein complex AP3, which is associated with coated vesicles
budding from the trans Golgi network. How might the loss of AP3 cause a
defect in melanosomes?
normal mouse
Mocha mouse
Figure 13–10 A normal mouse and
the Mocha mouse (Problem 13–60). In
addition to its light coat color, the Mocha
mouse has a poor sense of balance.
TRANSPORT FROM THE TRANS GOLGI NETWORK TO LYSOSOMES
(A)
(B)
normal and Ashen mice
Figure 13–11 Pigmentation defects in
Ashen mice (Problem 13–61). (A) Normally
pigmented mice and pale Ashen mice.
(B) A melanocyte from a normal mouse.
(C) Melanocytes from an Ashen mouse.
(C)
normal melanocyte
Ashen melanocytes
DATA HANDLING
13–61
271
More than 50 diferent genes are known to afect coat color in mice. hree
of them—Dilute, Leaden, and Ashen—are grouped together because of
their highly similar phenotypes. Although these mice have normal melanosomes in their melanocytes, the pigment in the melanosomes is not
delivered correctly to hair cells, giving rise to pale coats, as shown for
Ashen mice in Figure 13–11A. Dilute mice lack an unconventional myosin heavy chain, MyoVa, which interacts with a microtubule-based transport motor. Ashen mice carry a mutation in the gene for the Rab protein,
Rab27a, which associates with melanosomes. Leaden mice are missing melanophilin (Mlph), which is a modular protein with individual
domains that bind to MyoVa, to Rab27a, and to actin ilaments in the cell
cortex.
Melanocytes from normal mice have a characteristic branch morphology (Figure 13–11B) and normally discharge their melanosomes
near the tips of the branches. As shown in Figure 13–11C, melanocytes
from Ashen mice have a normal morphology but their melanosomes surround the nucleus. Melanocytes from Dilute mice and Leaden mice have
the same appearance as those from Ashen mice. Try to put these observations together to formulate a hypothesis to account for the normal delivery of melanosomes to the tips of the melanocyte branches, and for the
defects in melanosome function in these mice.
MEDICAL LINKS
13–62
Patients with I-cell disease are missing the enzyme GlcNAc phosphotransferase, which catalyzes the irst of the two steps required for addition of phosphate to mannose to create the M6P marker (see Figure
13–12). In the absence of the M6P marker, the M6P receptor cannot bind
to the protein and deliver it to a lysosome. How do you suppose that the
lysosomes in some cells from these patients—liver cells, for example—
acquire a normal complement of lysosomal enzymes?
13–63
Patients with Hunter’s syndrome or with Hurler’s syndrome rarely live
beyond their teens. hese patients accumulate glycosaminoglycans in
lysosomes due to the lack of speciic lysosomal enzymes necessary for
their degradation. When cells from patients with the two syndromes are
fused, glycosaminoglycans are degraded properly, indicating that the
cells are missing diferent degradative enzymes. Even if the cells are just
cultured together, they still correct each other’s defects. Most surprising
of all, the medium from a culture of Hurler’s cells corrects the defect in
Hunter’s cells (and vice versa). he corrective factors in the media are
inactivated by treatment with proteases, by treatment with periodate,
which destroys carbohydrate, and by treatment with alkaline phosphatase, which removes phosphates.
Chapter 13: Intracellular Membrane Traffic
272
UDP
Figure 13–12 Synthesis of M6P marker on
a lysosomal hydrolase (Problem 13–64).
UMP
GlcNAc
GlcNAc
CH2OH
CH2O
OH
CH2O
GlcNAc
O
O
OH
OH
O
OH
P
GlcNAc
phosphotransferase
OH
α -D-mannose
O
OH
OH
O
P
GlcNAc
phosphoglycosidase
OH
OH
mannose 6-phosphate
A. What do you suppose the corrective factors are? Beginning with the
donor patient’s cells, describe the route by which the factors reach the
medium and subsequently enter the recipient cells to correct the lysosomal defects.
B. Why do you suppose the treatments with protease, periodate, and alkaline phosphatase inactivate the corrective factors?
C. Would you expect a similar sort of correction scheme to work for mutant
cytosolic enzymes?
13–64
Children with I-cell disease synthesize perfectly good lysosomal
enzymes, but they are secreted outside the cell instead of being sorted
to lysosomes. he mistake occurs because the cells lack GlcNAc phosphotransferase, which is required to create the M6P marker that is essential for proper sorting (Figure 13–12). In principle, I-cell disease could
also be caused by deiciencies in GlcNAc phosphoglycosidase, which
removes GlcNAc to expose M6P (Figure 13–12), or in the M6P receptor
itself. hus, there are three potential kinds of I-cell disease, which could
be distinguished by the ability of various culture supernatants to correct
defects in mutant cells. Imagine that you have three cell lines (A, B, and
C), each of which derives from a patient with one of the three hypothetical I-cell diseases. Experiments with supernatants from these cell lines
give the results below.
1. he supernatant from normal cells corrects the defects in B and C but
not the defect in A.
2. he supernatant from A corrects the defect in Hurler’s cells, which are
missing a speciic lysosomal enzyme, but the supernatants from B and C
do not.
3. If the supernatants from the mutant cells are irst treated with phosphoglycosidase to remove GlcNAc, then the supernatants from A and C
correct the defect in Hurler’s cells, but the supernatant from B does not.
From these results, deduce the nature of the defect in each of the
mutant cell lines.
TRANSPORT INTO THE CELL FROM THE PLASMA
MEMBRANE: ENDOCYTOSIS
TERMS TO LEARN
caveola
caveolin
clathrin-coated pit
early endosome
endocytic vesicle
endocytosis
endosome maturation
ESCRT protein complexes
late endosome
low-density lipoprotein (LDL)
macrophage
O
macropinocytosis
multivesicular body
neutrophil
phagocytosis
phagosome
pinocytosis
receptor-mediated endocytosis
recycling endosome
transcytosis
transferrin receptor
TRANSPORT INTO THE CELL FROM THE PLASMA MEMBRANE: ENDOCYTOSIS
DEFINITIONS
Match each deinition below with its term from the list above.
13–65
General term for the process by which cells take up macromolecules,
particulate substances, and even other cells into membrane-enclosed
vesicles.
13–66
Complex vesicle with invaginating buds and internal vesicles involved in
the maturation of early endosomes into late endosomes.
13–67
Phagocytic cell—derived from a hematopoietic stem cell—that ingests
invading microorganisms and plays an important role in scavenging
senescent cells and apoptotic cells.
13–68
Type of endocytosis in which soluble materials are taken up from the
environment and incorporated into vesicles for digestion.
13–69
Invagination that forms from lipid rafts at the cell surface and buds of
internally to form a pinocytic vesicle.
13–70
Region of plasma membrane of animal cells that is covered with the protein clathrin on its cytosolic face; it will bud of from the membrane to
form an intracellular vesicle.
13–71
Process by which macromolecules bind to complementary transmembrane receptor proteins, accumulate in coated pits, and then enter the
cells as receptor–macromolecule complexes in clathrin-coated vesicles.
13–72
One of a family of structural proteins in caveolae that are unusual because
they extend multiple hydrophobic loops into the membrane from the
cytosolic side, but do not cross the membrane.
13–73
Membrane-enclosed compartment just beneath the plasma membrane,
to which external molecules are irst delivered by endocytosis.
13–74
Specialized form of endocytosis in which a cell uses large endocytic vesicles to ingest large particles such as microorganisms and dead cells.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
13–75
Any particle that is bound to the surface of a phagocyte will be ingested
by phagocytosis.
13–76
Like the LDL receptor, most of the more than 25 diferent receptors
known to participate in receptor-mediated endocytosis enter coated pits
only after they have bound their speciic ligands.
13–77
All the molecules that enter early endosomes ultimately reach late
endosomes, where they become mixed with newly synthesized acid
hydrolases and end up in lysosomes.
13–78
During transcytosis, vesicles that form from coated pits on the apical surface fuse with the plasma membrane on the basolateral surface, and in
that way transport molecules across the epithelium.
THOUGHT PROBLEMS
13–79
A macrophage ingests the equivalent of 100% of its plasma membrane
each half hour by endocytosis. What is the rate at which membrane is
returned by exocytosis?
273
Chapter 13: Intracellular Membrane Traffic
274
(A) MICROGRAPH
Figure 13–13 A coated pit about to bud
from the membrane (Problem 13–80).
(A) An electron micrograph.
(B) A schematic drawing.
(B) DRAWING
C
A
B
G
F
D
E
100 nm
he electron micrograph in Figure 13–13A is illustrated schematically by
the drawing in Figure 13–13B. Name the structures that are labeled in the
drawing.
13–81
Caveolae are thought to form from lipid rafts, which are patches of the
plasma membrane that are especially rich in cholesterol and glycosphingolipids. Caveolae may collect cargo proteins by virtue of the lipid composition of their membrane, rather than by assembly of a cytosolic protein
coat. What might you predict would be a characteristic of the structure of
transmembrane proteins that collect in caveolae?
13–82
Iron (Fe) is an essential trace metal that is needed by all cells. It is
required, for example, for the synthesis of the heme groups that are part
of cytochromes and hemoglobin. Iron is taken into cells via a two-component system. he soluble protein transferrin circulates in the bloodstream, and the transferrin receptor is a membrane protein that is continually endocytosed and recycled to the plasma membrane. Fe ions
bind to transferrin at neutral pH but not at acidic pH. Transferrin binds
to the transferrin receptor at neutral pH only when it has bound an Fe
ion, but it binds to the receptor at acidic pH even in the absence of bound
iron. From these properties, describe how iron is taken up, and discuss
the advantages of this elaborate scheme.
CALCULATIONS
Cells take up extracellular molecules by receptor-mediated endocytosis
and by luid-phase endocytosis. A classic paper compared the eiciencies of these two pathways by incubating human cells for various periods of time in a range of concentrations of either 125I-labeled epidermal
growth factor (EGF), to measure receptor-mediated endocytosis, or
horseradish peroxidase (HRP), to measure luid-phase endocytosis. Both
EGF and HRP were found to be present in small vesicles with an internal
radius of 20 nm. he uptake of HRP was linear (Figure 13–14A), while
that of EGF was initially linear but reached a plateau at higher concentrations (Figure 13–14B).
A. Explain why the shapes of the curves in Figure 13–14 are diferent for
HRP and EGF.
B. From the curves in Figure 13–14, estimate the diference in the uptake
rates for HRP and EGF when both are present at 40 nM. What would the
diference be if both were present at 40 μM?
C. Calculate the average number of HRP molecules that get taken up by
each endocytic vesicle (radius 20 nm) when the medium contains 40 μM
HRP. [he volume of a sphere is (4/3)πr3.]
(A) HRP UPTAKE
HRP uptake
(pmol/hour/106 cells)
13–80
2
1
0
0
13–83
10
20
30
40
HRP in medium (μM)
(B) EGF UPTAKE
125I-EGF uptake
(pmol/hour/106 cells)
20
16
12
8
4
0
0
20
40
60
80
EGF in medium (nM)
Figure 13–14 Uptake of HRP and EGF
as a function of their concentration in the
medium (Problem 13–83).
TRANSPORT INTO THE CELL FROM THE PLASMA MEMBRANE: ENDOCYTOSIS
275
D. he scientists who did these experiments said at the time, “hese calculations clearly illustrate how cells can internalize EGF by endocytosis while
excluding all but insigniicant quantities of extracellular luid.” What do
you think they meant?
13–84
A ligand for receptor-mediated endocytosis circulates at a concentration
of 1 nM (10–9 M). It is taken up in coated vesicles with a volume of 1.66 ×
10–18 L (about 150 nm in diameter). On average, there are 10 of its receptors in each coated vesicle. If all the receptors were bound to the ligand,
how much more concentrated would the ligand be in the vesicle than it
was in the extracellular luid? What would the dissociation constant (Kd)
for the receptor–ligand binding need to be in order to concentrate the
ligand 1000-fold in the vesicle? (You may wish to review the discussion of
Kd in Problem 3–86.)
he recycling of transferrin receptors has been studied by labeling the
receptors on the cell surface and following their fate at 0°C and 37°C. A
sample of intact cells at 0°C was reacted with radioactive iodine under
conditions that label cell-surface proteins. If these cells were kept on
ice and incubated in the presence of trypsin, which destroys the receptors without damaging the integrity of the cell, the radioactive transferrin receptors were completely degraded. If the cells were irst warmed
to 37°C for 1 hour and then treated with trypsin on ice, about 70% of the
initial radioactivity was resistant to trypsin. At both temperatures, most
of the receptors were not labeled and most remained intact, as apparent
from a protein stain.
A second sample of cells that had been surface-labeled at 0°C and
incubated at 37°C for 1 hour was analyzed with transferrin-speciic antibodies, which identify transferrin receptors via their linkage to Fe–transferrin complexes. If intact cells were reacted with antibody, 0.54% of the
labeled proteins were bound by antibody. If the cells were irst dissolved
in detergent, 1.76% of the labeled proteins were bound by antibody.
A. When the cells were kept on ice, why did trypsin treatment destroy the
labeled transferrin receptors, but not the majority of receptors? Why did
most of the labeled receptors become resistant to trypsin when the cells
were incubated at 37°C?
B. What fraction of all the transferrin receptors is on the cell surface after a
1-hour incubation at 37°C? Do the two experimental approaches agree?
13–85
13–86
Cholesterol is an essential component of the plasma membrane, but
people who have very high levels of cholesterol in their blood (hypercholesterolemia) tend to have heart attacks. Blood cholesterol is carried
in the form of cholesterol esters in low-density lipoprotein (LDL) particles. LDL binds to a high-ainity receptor on the cell surface, enters the
cell via a coated pit, and ends up in lysosomes. here its protein coat is
degraded, and cholesterol esters are released and hydrolyzed to cholesterol. he released cholesterol enters the cytosol and inhibits the enzyme
HMG CoA reductase, which controls the irst unique step in cholesterol
biosynthesis. Patients with severe hypercholesterolemia cannot remove
LDL from the blood. As a result, their cells do not turn of normal cholesterol synthesis, which makes the problem worse.
LDL metabolism can be conveniently divided into three stages experimentally: binding of LDL to the cell surface, internalization of LDL,
and regulation of cholesterol synthesis by LDL. Skin cells from a normal
person and two patients sufering from severe familial hypercholesterolemia were grown in culture and tested for LDL binding, LDL internalization, and LDL regulation of cholesterol synthesis. he results are shown in
Figure 13–15.
LDL bound
(μg/g cell protein)
normal
normal
JD
0.1
FH
0.0
0
50
100
0
50
100
LDL internalized
(μ g/g cell protein)
(B) INTERNALIZATION
1.0
normal
normal
0.5
JD
FH
0.0
0
50
100
0
50
100
(C) REGULATION
cholesterol synthesis
(nmol/hr)
MEDICAL LINKS
(A) BINDING
0.2
0.4
FH
JD
normal
normal
0.2
0.0
0
50
100
LDL (μ g/mL)
0
50
100
LDL (μ g/mL)
Figure 13–15 lDl metabolism in normal
cells and in cells from patients with
severe familial hypercholesterolemia
(Problem 13–86). (A) Surface binding of
lDl. Assays at 4°C allow binding but not
internalization. (B) Internalization of lDl.
After binding at 4°C, the cells are warmed
to 37°C. Binding and uptake of lDl can
be followed by labeling lDl either with
ferritin particles, which can be seen by
electron microscopy, or with radioactive
iodine, which can be measured in
a gamma counter. (C) Regulation of
cholesterol synthesis by lDl.
Chapter 13: Intracellular Membrane Traffic
276
A. In Figure 13–15A, the surface binding of LDL by normal cells is compared
with LDL binding by cells from patients FH and JD. Why does binding by
normal cells and by JD’s cells reach a plateau? What explanation can you
suggest for the lack of LDL binding by FH’s cells?
B. In Figure 13–15B, internalization of LDL by normal cells increases as the
external LDL concentration is increased, reaching a plateau 5-fold higher
than the amount of externally bound LDL. Why does LDL enter cells from
patients FH or JD at such a slow rate?
C. In Figure 13–15C, the regulation of cholesterol synthesis by LDL in
normal cells is compared with that in cells from FH and JD. Why does
increasing the external LDL concentration inhibit cholesterol synthesis
in normal cells, but afect it only slightly in cells from FH or JD?
D. How would you expect the rate of cholesterol synthesis to be afected if
normal cells and cells from FH or JD were incubated with cholesterol
itself? (Free cholesterol crosses the plasma membrane by difusion.)
What is wrong with JD’s metabolism of LDL? As discussed in Problem
13–86, JD’s cells bind LDL with the same ainity as normal cells and in
almost the same amounts, but the binding does not lead to internalization of LDL. Two classes of explanation could account for JD’s problem:
1. JD’s LDL receptors are defective in a way that prevents internalization, even though the LDL-binding domains on the cell surface are unaffected.
2. JD’s LDL receptors are entirely normal, but there is a mutation in the
cellular internalization machinery such that loaded LDL receptors cannot be brought in.
To distinguish between these explanations, JD’s parents were studied. It is known that an autosomal gene encodes the receptor that binds
LDL. hus, each parent must have donated one defective gene to JD. JD’s
mother sufered from mildly elevated blood cholesterol. Her cells bound
only half as much LDL as normal cells, but the bound LDL was internalized at the same rate as in normal cells. JD’s father also had mild hypercholesterolemia, but his cells bound even more LDL than normal cells.
Of the bound LDL, less than half the label could be internalized; the rest
remained on the cell surface.
he association of this family’s LDL receptors with coated pits was
studied by electron microscopy (EM), using LDL that was labeled with
ferritin. he results are shown in Table 13–5.
A. Why does JD’s mother have mild hypercholesterolemia? Based on the
LDL binding and internalization studies, and on the EM observations,
decide what kind of defective LDL receptor gene she passed to JD.
13–87
TABLE 13–5 Distribution of LDL receptors on the surface of cells from
JD and his parents as compared with normal individuals (Problem
13–87).
Number of lDl receptors
Individual
In pits
outside pits
Normal male
186
195
Normal female
186
165
JD
10
342
JD’s father
112
444
JD’s mother
91
87
TRANSPORT FROM THE TRANS GOLGI NETWORK TO THE CELL EXTERIOR: EXOCYTOSIS
B. Why does JD’s father have mild hypercholesterolemia? Based on the LDL
binding and internalization studies, and on the EM observations, decide
what kind of defective LDL receptor gene he passed to JD.
C. Can you account for JD’s hypercholesterolemia from the behavior of the
LDL receptors in his parents? In particular, how is it that JD binds nearly
a normal amount of LDL, but has severe hypercholesterolemia?
D. At the beginning of this problem, two possible explanations—defective receptor or defective internalization machinery—were proposed to
account for the lack of internalization by JD’s LDL receptors in the face
of nearly normal LDL binding. Do these studies allow you to decide
between these alternative explanations?
TRANSPORT FROM THE TRANS GOLGI NETWORK
TO THE CELL EXTERIOR: EXOCYTOSIS
TERMS TO LEARN
constitutive secretory pathway
default pathway
exocytosis
regulated secretory pathway
secretory vesicle
synaptic vesicle
DEFINITIONS
Match each deinition below with its term from the list above.
13–88
Specialized class of tiny secretory vesicles that store neurotransmitter
molecules.
13–89
Pathway for exocytosis that operates continuously in all cells.
13–90
Membrane-enclosed organelle in which molecules destined to be
exported are stored prior to release.
13–91
Process involving fusion of vesicles with the plasma membrane.
13–92
Pathway for exocytosis that operates mainly in cells specialized for secreting products rapidly on demand.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
13–93
When a foreign gene encoding a secretory protein is introduced into a
secretory cell that normally does not make the protein, the alien secretory protein is not packaged into secretory vesicles.
13–94
Once a secretory vesicle is properly positioned beneath the plasma
membrane, it will immediately fuse with the membrane and release its
contents to the cell exterior.
THOUGHT PROBLEMS
13–95
In a cell capable of regulated secretion, what are the three main classes of
protein that must be separated before they leave the trans Golgi network?
13–96
You are interested in exocytosis and endocytosis in a line of cultured liver
cells that secrete albumin and take up transferrin. To distinguish between
these events, you tag transferrin with colloidal gold and prepare ferritinlabeled antibodies that are speciic for albumin. You add the tagged
transferrin to the medium, and then after a few minutes you ix the cells,
prepare thin sections, and react them with ferritin-labeled antibodies
against albumin. Colloidal gold and ferritin are both electron-dense and
277
Chapter 13: Intracellular Membrane Traffic
278
therefore readily visible when viewed by electron microscopy; moreover,
they can be easily distinguished from one another on the basis of size.
A. Will this experiment allow you to identify vesicles in the exocytic and
endocytic pathways? How?
B. Not all the gold-labeled vesicles are clathrin-coated. Why?
13–97
13–98
What would you expect to happen in cells that secrete large amounts of
protein through the regulated secretory pathway, if the ionic conditions
in the ER lumen could be changed to resemble those of the trans Golgi
network?
Dynamin was irst identiied as a microtubule-binding protein, and its
sequence indicated that it was a GTPase. he key to its function came
from neurobiological studies in Drosophila. Shibire mutant lies, which
carry a mutation in the dynamin gene, are rapidly paralyzed when the
temperature is elevated. hey recover quickly once the temperature is
lowered. he complete paralysis at the elevated temperature suggested
that synaptic transmission between nerve and muscle cells was blocked.
Electron micrographs of synapses of the paralyzed lies showed a loss of
synaptic vesicles and a tremendously increased number of coated pits
relative to normal synapses (Figure 13–16).
Suggest an explanation for the paralysis shown by the shibire mutant
lies, and indicate why signal transmission at a synapse might require
dynamin.
DATA HANDLING
13–99
Proteins without special signals are transported between cisternae in
Golgi stacks and onward to the plasma membrane via the nonselective
constitutive secretory pathway, or the default pathway, as it is commonly
known. his transport is also sometimes referred to as bulk transport
because the Golgi contents do not become concentrated in the vesicles.
Given that transport in clathrin-coated vesicles is so highly concentrating, you are skeptical that no concentration occurs in the default pathway
for secretion.
To determine whether vesicles in the default pathway concentrate
their contents, you infect cells with vesicular stomatitis virus (VSV) and
follow the viral G protein, which is transported by the default pathway.
Your idea, an ambitious one, is to compare the concentration of G protein in the lumen of the Golgi stacks with that in the associated transport
vesicles. You intend to measure G-protein concentration by preparing
thin sections of VSV-infected cells and incubating them with G-proteinspeciic antibodies tagged with gold particles. Since the gold particles are
visible in electron micrographs as small black dots, it is relatively straightforward to count dots in the lumens of transport vesicles (fully formed
and just budding) and of the Golgi apparatus. You make two estimates
of G-protein concentration: (1) the number of gold particles per crosssectional area (surface density) and (2) the number of gold particles per
linear length of membrane (linear density). Your results are shown in
Table 13–6.
Do the vesicles involved in the default pathway concentrate their contents or not? Explain your reasoning.
13–100 Insulin is synthesized as a pre-pro-protein in the β cells of the pancreas.
Its pre-peptide is cleaved of after it enters the ER lumen. To deine the
cellular location at which its pro-peptide is removed, you have prepared
two antibodies: one that is speciic for pro-insulin, and one that is speciic for insulin. You have tagged the anti-pro-insulin antibody with a red
luorophore and the anti-insulin antibody with a green luorophore, so
you can follow them independently in the same cell. When you incubate
extracellular space
cytosol
100 nm
Figure 13–16 Electron micrograph of a
nerve terminal from a shibire mutant ly at
elevated temperature (Problem 13–98).
TRANSPORT FROM THE TRANS GOLGI NETWORK TO THE CELL EXTERIOR: EXOCYTOSIS
TABLE 13–6 Relative densities of G protein in Golgi and vesicle lumens
and membranes (Problem 13–99).
Source of Golgi
Site measured
Parameter
measured
Mean
density
Uninfected cells
Whole Golgi
Surface density
5/μm2
Infected cells
Whole Golgi
Surface density
271/μm2
Infected cells
Golgi buds and
vesicles
Surface density
233/μm2
Infected cells
Golgi cisternal
membranes
Linear density
6/μm
Infected cells
Golgi buds and
vesicles
Linear density
4/μm
a pancreatic β cell with a mixture of your two antibodies, you obtain the
results shown in Table 13–7. In what cellular compartment is the propeptide removed from pro-insulin?
13–101 Polarized epithelial cells must make an extra sorting decision since their
plasma membranes are divided into apical and basolateral domains,
which are populated by distinctive sets of proteins. Proteins destined for
the apical or basolateral domains seem to travel there directly from the
trans Golgi network. One way to sort proteins to these domains would
be to use a speciic sorting signal for one group of proteins, which would
then be actively recognized and directed to one domain, and to allow the
rest to travel via a default pathway to the other domain.
Consider the following experiment to identify the default pathway.
he cloned genes for several foreign proteins were engineered by recombinant DNA techniques so that they could be expressed in the polarized
epithelial cell line MDCK. hese proteins are secreted in other types of
cells, but are not normally expressed in MDCK cells. he cloned genes
were introduced into the polarized MDCK cells, and their sites of secretion were assayed. Although the cells remained polarized, the foreign
proteins were delivered in roughly equal amounts to the apical and basolateral domains.
TABLE 13–7 Fluorescence associated with various compartments
of β cells after reaction with fluorescent antibodies directed against
pro-insulin and insulin (Problem 13–100).
Compartment
Fluorescence
cis Golgi network
Red
Endoplasmic reticulum
Red
Golgi cisternae
Red
Immature secretory vesicles
Yellow
Lysosomes
None
Mature secretory vesicles
Green
Mitochondria
None
Nucleus
None
trans Golgi network
Red
279
280
Chapter 13: Intracellular Membrane Traffic
A. What is the expected result of this experiment, based on the hypothesis
that targeting to one domain of the plasma membrane is actively signaled
and targeting to the other domain is via a default pathway?
B. Do these results support the concept of a default pathway as outlined
above?
13–102 Neurons are diicult to study because of their excessively branched struc-
ture and long, thin dendrites, as shown in Figure 13–17. Fluorescently
tagged antibodies are powerful tools for investigating certain aspects of
neuron structure. Synaptic vesicles, for example, were shown to be concentrated in the presynaptic cells at nerve synapses in this way. A culture
of neurons was irst exposed for 1 hour to a luorescently tagged antibody
speciic for the lumenal domain of synaptotagmin, a transmembrane
protein that resides exclusively in the membranes of synaptic vesicles.
he culture was then washed thoroughly to remove all synaptotagmin
antibodies. When the culture was examined by luorescence microscopy,
dots of color from the synaptotagmin-speciic antibody were found to
mark the positions of the synaptic vesicles in the nerve terminals.
If antibodies do not cross intact membranes, how do you suppose the
synaptic vesicles get labeled? When the procedure was repeated using
an antibody speciic for the cytoplasmic domain of synaptotagmin, the
nerve terminals did not become labeled. Explain the results with the two
diferent antibodies for synaptotagmin.
Figure 13–17 A hippocampal neuron
(Problem 13–102).
13–103 he original version of the SNARE hypothesis suggested that the ATP-
dependent disassembly of SNAREs by NSF provided the energy necessary for membrane fusion, and thus that NSF should act at the last step in
secretion. More recent evidence suggests that NSF acts at an earlier step
to prime the vesicle for secretion, and that SNAREs alone are suicient
to catalyze membrane fusion at the last step in secretion. It is important
to know what really happens, and you have the means at hand to answer
the question.
Using a whole-cell patch-clamp protocol (Figure 13–18A), you
can difuse cytosolic components into the cell through the pipet and
assay exocytosis by changes in capacitance, which is a measure of the
increase in the area of the plasma membrane. To control precisely the
timing of vesicle fusion, you enclose Ca2+ in a photosensitive chemical
“cage,” from which it can be released with a lash of light. In response to
Ca2+ release, there is a rapid burst of vesicle fusion (indicated by a
rapid rise in capacitance), followed by a longer, slower fusion process
(Figure 13–18B, –NEM). he initial burst represents the fusion of vesicles that were just waiting for the Ca2+ trigger. Because Ca2+ is rapidly
removed from the cell, the procedure can be repeated with a second lash
of light 2 minutes later; it yields the same rapid and slow components
(Figure 13–18C, –NEM).
(A)
(B)
Figure 13–18 Analysis of the role of NSF in
vesicle fusion (Problem 13–103).
(A) Whole-cell patch-clamp. (B) Responses
to the irst lash-mediated release of
Ca2+. (C) reponses to the second lashmediated release of Ca2+. secretion was
measured as an increase in membrane
capacitance (in femtoFarads, f F) in the
absence (–neM) or presence (+neM) of an
inhibitor of nsF.
(C)
FIRST FLASH
SECOND FLASH
– NEM
+ NEM
200 f F
+ NEM
200 f F
patch
pipet
membrane capacitance
(secretion)
− NEM
0
cell
5
time (seconds)
10
0
5
time (seconds)
10
MCAT STYLE
To test the role of NSF in vesicle fusion, you irst difuse N-ethylmaleimide (NEM) into cells to inhibit NSF. (he name “NSF” stands for
NEM-sensitive factor.) You then repeat the lash protocols as before. In
response to the irst lash, the rapid component is unafected, but the
slow component is decreased (Figure 13–18B, +NEM). In response to the
second lash, both components are inhibited (Figure 13–18C, +NEM).
A. What do you suppose the slow component of the fusion process represents?
B. Why does inhibition of NSF afect the slow component after both lashes,
but inhibit the rapid component only after the second lash?
C. Which of the alternatives for the role of NSF in vesicle fusion—acting at
the last step or an early step—do these experiments support? Explain
your reasoning.
D. Propose a model for the molecular role of NSF in fusion of secretory
vesicles.
MEDICAL LINKS
13–104 Antitrypsin, which inhibits certain proteases, is normally secreted into the
bloodstream by liver cells. Antitrypsin is absent from the bloodstream of
patients who carry a mutation that results in a single amino acid change
in the protein. Antitrypsin deiciency causes a variety of severe problems,
particularly in lung tissue, because of uncontrolled protease activity. Surprisingly, when the mutant antitrypsin is synthesized in the laboratory,
it is as active as the normal antitrypsin at inhibiting proteases. Why then
does the mutation cause the disease? hink of more than one possibility
and suggest ways in which you could distinguish among them.
MCAT STYLE
Passage 1 (Questions 13–105 to 13–107)
Ricin is one of the most toxic substances known: less than 2 mg injected into the
bloodstream will kill an adult human. Ricin is produced by the castor bean plant
as a 65 kd protein heterodimer composed of an A chain and a B chain. he B chain
is a lectin that binds to carbohydrates on the cell surface. he A chain is an enzyme
that modiies a highly conserved site in rRNA, leading to inhibition of translation.
After entering the cell, ricin eventually ends up in the lumen of the endoplasmic
reticulum (ER), and from there it moves into the cytosol, where it inactivates ribosomes.
13–105 What is the most likely mechanism by which ricin enters the cell?
A. Binding to clathrin proteins
B. Entry through pore complexes
C. Interaction with SNARE proteins
D. Internalization via endocytosis
13–106 Which one of the following is required in order for ricin to be delivered to
the ER?
A. N-ethylmaleimide-sensitive factor (NSF)
B. Golgi-derived COPI-coated vesicles
C. Mannose 6-phosphate (M6P) receptors
D. he Sar1 monomeric GTPase
13–107 Which one of the following describes the most likely scenario for how
ricin gets into the cytosol?
A. Ricin has a signal sequence that allows it to be transported across the ER
membrane into the cytosol.
B. Ricin is packaged into vesicles that form vesicular tubular clusters, which
release proteins into the cytosol.
281
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Chapter 13: Intracellular Membrane Traffic
C. Ricin is transported from the ER to the Golgi apparatus to the lysosome,
where it is released into the cytosol.
D. Ricin, by mimicking an unfolded protein, is tagged for transport across
the ER membrane into the cytosol.
Passage 2 (Questions 13–108 to 13–111)
he molecular machinery responsible for vesicle fusion was discovered via biochemical analysis of vesicle transport. hese experiments used a virus that hijacks
the host-cell machinery to insert a viral coat protein into the ER membrane. From
there, the viral protein is transported through the Golgi apparatus to the cell surface, where it is packaged into new virus particles that bud of the cell surface. In
the mature virus, the viral protein is exposed to the outside of the virus and forms
part of the viral coat. As the viral protein moves through the Golgi apparatus, it is
modiied with N-acetylglucosamine (GlcNAc).
he investigators used the virus to infect a mutant cell line that is incapable of
modifying the viral protein with GlcNAc. he Golgi apparatus was then isolated
from the infected cells. his “donor” Golgi apparatus was mixed with “acceptor” Golgi apparatus isolated from uninfected wild-type cells. he investigators
hypothesized that transport of vesicles from the donor to acceptor Golgi apparatuses would move the viral protein into the wild-type acceptor Golgi apparatus,
leading to addition of GlcNAc. hus, vesicle traic between Golgi stacks could be
detected simply by assaying for the GlcNAc modiication on the viral protein.
In addition to purifying donor and acceptor Golgi apparatuses, the investigators made a preparation of soluble cytosolic proteins by breaking open cells and
removing membrane-bound organelles and other large particles by centrifugation. When the donor and acceptor Golgi apparatuses were mixed in the absence
of cytosol, no transport of the viral protein could be detected. However, when ATP
and soluble cytosolic proteins were included, the investigators detected transport of viral protein between the Golgi stacks. To identify proteins required for
vesicle transport, the investigators treated the soluble cytoplasmic extract with
N-ethylmaleimide (NEM), which covalently modiies lysine residues, inactivating some proteins. his process inactivated the activity in the extract, which
allowed the investigators to purify a protein from untreated extracts that could be
added back to NEM-treated extract to rescue vesicle transport. his protein was
called NEM-sensitive factor (NSF). NSF was found to be a soluble protein that can
hydrolyze ATP.
NSF was found to bind to the membranes of the Golgi apparatus in the presence
of soluble proteins called “soluble NSF attachment proteins” (SNAPs). Complexes
of NSF and SNAP formed only in the absence of ATP; addition of ATP caused rapid
release of NSF. If Golgi membranes with bound NSF were solubilized by addition
of nonionic detergent, NSF was found to be associated with a large protein complex. To identify proteins that bind to NSF, the investigators attached an antibody
that recognizes NSF to beads. he beads were then used to bind detergent-solubilized NSF—with its associated complex of proteins—from bovine brain extracts.
After washing with bufer to remove unbound proteins, ATP was added to release
the proteins that bind to NSF and SNAP. hese proteins were called SNAP receptors (SNAREs). Two of the proteins the investigators identiied—syntaxin and
synaptobrevin—had already been found in other studies of neuronal cell exocytosis. SNAREs were hypothesized to be the minimal machinery necessary for
membrane fusion events in the secretory pathway. To test this, a v-SNARE and a
t-SNARE were expressed in bacteria, puriied, and reconstituted into lipid vesicles. When vesicles bearing the v-SNARE were mixed with those containing the
t-SNARE, the vesicles fused. NSF and ATP were not required for fusion.
13–108 he investigators hypothesized that the viral protein was transported
between the Golgi stacks inside vesicles. An alternative hypothesis, however, was that the viral protein was released from one Golgi apparatus
and taken up by the other, without being packaged into vesicles. Which
MCAT STYLE
of the following experiments would best distinguish between these two
hypotheses?
A. Add a protease to the system and determine whether the viral protein is
degraded.
B. Determine whether transport still occurs when clathrin is removed from
the extract.
C. Test for association of GlcNAc-modiied viral protein with membranes by
centrifugation.
D. Test whether transport between Golgi stacks is blocked by addition of a
detergent.
13–109 To learn more about the function of NSF, the investigators used a cyto-
plasmic extract that lacked NSF activity. To the extract, they added donor
Golgi apparatus, puriied NSF, and ATP. hey then added NEM to inactivate NSF, followed by acceptor membranes. No vesicle transport was
detected in this situation. What does this experiment tell you about the
function of NSF?
A. NSF is required for packaging viral protein into donor vesicles.
B. NSF is required for formation of donor vesicles from donor Golgi.
C. NSF is required for fusion of donor vesicles to the acceptor Golgi.
D. NSF is required for release of viral protein into the acceptor Golgi.
13–110 Which of the following statements about synaptobrevin and syntaxin are
consistent with what we now know about the key functions of SNAREs?
I. Botulinum toxin proteolytically cleaves synaptobrevin, leading to muscle
paralysis.
II. Synaptobrevin induces NSF to hydrolyze ATP to provide the energy for
membrane fusion.
III. Synaptobrevin is on the vesicle membrane; syntaxin is on the plasma
membrane.
A. I
B. I and II
C. I and III
D. II and III
13–111 Which one of the following statements best explains why NSF and ATP
were required for vesicle fusion in the original reconstituted system for
transport of a viral protein between Golgi stacks, but were not required
for fusion of lipid vesicles containing a v-SNARE and a t-SNARE?
A. Fusion of lipid vesicles requires no extra energy in the presence of high
concentrations of puriied SNAREs; however, energy from ATP hydrolysis
is required to fuse vesicles with natural membranes.
B. In the fusion of natural membranes, NSF and ATP are required to remove
a cytosolic inhibitor that binds to SNARE proteins; the fusion of lipid vesicles occurs without ATP because the inhibitor is absent.
C. NSF uses ATP hydrolysis to move proteins out of the way, so that natural
vesicles can dock on their target membrane, but ATP hydrolysis is not
needed to dock lipid vesicles containing puriied SNAREs.
D. NSF uses ATP to pull apart tight complexes between v-SNAREs and
t-SNAREs to prime them for fusion, a process that is not required when
the SNAREs are already segregated into diferent lipid vesicles.
283
High-Current Copper-Brush
Commutated Dynamo.
According to Wikipedia, this large,
belt-driven, high-current dynamo
produced 310 amperes at 7 volts.
Mitochondria are often referred to as the
“power plants” of the cell. The pictured
machine converted mechanical energy
into electrical energy; mitochondria
convert electrical energy into chemical
energy.
Chapter 14
285
CHAPTER
Energy Conversion:
Mitochondria and Chloroplasts
14
THE MITOCHONDRION
IN THIS CHAPTER
TERMS TO LEARN
chemiosmotic coupling
cristae
electrochemical gradient
inner mitochondrial membrane
intermembrane space
mitochondrial matrix
THE MITOCHONDRION
mitochondria
outer mitochondrial membrane
oxidative phosphorylation
proton-motive force
respiratory chain
DEFINITIONS
Match each deinition below with its term from the list above.
14–1
he subcompartment formed between the inner and outer mitochondrial membranes.
14–2
he mitochondrial electron-transport chain, which generates the proton
gradient across the inner mitochondrial membrane that powers ATP synthase, producing most of the cell’s ATP.
14–3
Mechanism by which a pH gradient across a membrane is used to drive
an energy-requiring process, such as ATP production or the rotation of
bacterial lagella.
14–4
Process in bacteria and mitochondria in which ATP formation is driven
by the transfer of electrons from food molecules to molecular oxygen,
with the intermediate generation of a proton gradient across a membrane.
14–5
he result of a combined pH gradient and membrane potential.
14–6
A sievelike membrane surrounding mitochondria that is permeable to all
molecules of 5000 daltons or less.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
14–7
he intermembrane space is chemically equivalent to the cytosol with
respect to small molecules due to the many specialized transport proteins in the mitochondrial outer membrane.
14–8
he most important contribution of the citric acid cycle to energy metabolism is the extraction of high-energy electrons during the oxidation of
acetyl CoA to CO2.
14–9
Each respiratory enzyme complex in the electron-transport chain has a
greater ainity for electrons than its predecessors, so that electrons pass
THE PROTON PUMPS OF THE
ELECTRON-TRANSPORT CHAIN
ATP PRODUCTION IN
MITOCHONDRIA
CHLOROPLASTS AND
PHOTOSYNTHESIS
THE GENETIC SYSTEMS
OF MITOCHONDRIA AND
CHLOROPLASTS
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
286
sequentially from one complex to another until they are inally transferred to oxygen, which has the greatest electron ainity of all.
THOUGHT PROBLEMS
14–10
Mitochondria in liver cells appear to move freely in the cytosol, whereas
those in cardiac muscle are immobilized at positions between adjacent
myoibrils. Do you suppose these diferences are a trivial consequence
of cell architecture or do they relect some underlying functional advantage? Explain your answer.
14–11
Electron micrographs show that mitochondria in heart muscle have a
much higher density of cristae than mitochondria in skin cells. Why do
you suppose this should be?
14–12
In the 1860s, Louis Pasteur noticed that when he added O2 to a culture of
yeast growing anaerobically on glucose, the rate of glucose consumption
declined dramatically. Explain the basis for this result, which is known as
the Pasteur efect.
14–13
he citric acid cycle generates NADH and FADH2, which are then used
in the process of oxidative phosphorylation to make ATP. If the citric acid
cycle, which does not use oxygen, and oxidative phosphorylation are
separate processes, as they are, then why is it that the citric acid cycle
stops almost immediately upon removal of O2?
When dinitrophenol (DNP) is added to mitochondria, the inner membrane becomes permeable to protons. When the drug valinomycin is
added to mitochondria, the inner membrane becomes permeable to K+.
A. How will the electrochemical gradient change in response to DNP?
B. How will it change in response to valinomycin?
14–14
14–15
Several coupled transport processes that occur across the inner mitochondrial membrane are illustrated in Figure 14–1. For each, decide
whether transport is with the electrochemical gradient, against it, or
unafected by it. For those transport processes that are afected by the
gradient, identify which component of the gradient (the diference in pH
or the membrane potential) afects transport.
CALCULATIONS
14–16
Heart muscle gets most of the ATP needed to power its continual contractions through oxidative phosphorylation. When oxidizing glucose to
CO2, heart muscle consumes O2 at a rate of 10 μmol/min per g of tissue,
in order to replace the ATP used in contraction and give a steady-state
ATP concentration of 5 μmol/g of tissue. At this rate, how many seconds would it take the heart to consume an amount of ATP equal to its
steady-state levels? (Complete oxidation of one molecule of glucose to
CO2 yields 30 ATP, 26 of which are derived by oxidative phosphorylation
using the 12 pairs of electrons captured in the electron carriers NADH
and FADH2.)
MATRIX
PYR–
I
H+
ORN+
II
CTR
III
HPO42–
2H+
CIT3– + H+
IV
MAL3–
Asp–
V
Figure 14–1 Five coupled transport processes that occur across the
inner mitochondrial membrane (Problem 14–15). PyR is pyruvate; oRN is
ornithine; CTR is citrulline; CIT is citrate; MAl is malate; Asp is aspartic
acid; and Glu is glutamic acid.
Glu–
inner
membrane
THE PROTON PUMPS OF THE ELECTRON-TRANSPORT CHAIN
THE PROTON PUMPS OF THE ELECTRONTRANSPORT CHAIN
TERMS TO LEARN
cytochrome
cytochrome c oxidase complex
cytochrome c reductase
iron–sulfur center
NADH dehydrogenase complex
quinone (Q)
redox pairs
redox potential
redox reaction
DEFINITIONS
Match each deinition below with its term from the list above.
14–17
An electron-driven proton pump in the respiratory chain that accepts
electrons from cytochrome c and generates water using molecular oxygen as an electron acceptor.
14–18
Colored, heme-containing protein that transfers electrons during cellular respiration.
14–19
Electron-transporting group consisting of either two or four iron atoms
bound to an equal number of sulfur atoms.
14–20
A reaction in which one component becomes oxidized and the other
reduced.
14–21
he ainity of a redox pair for electrons, generally measured as the voltage diference between an equimolar mixture of the pair and a standard
reference.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
14–22
Most cytochromes have a higher redox potential (higher ainity for electrons) than iron–sulfur centers, which is why the cytochromes tend to
serve as electron carriers near the O2 end of the respiratory chain.
14–23
he three respiratory enzyme complexes in the mitochondrial inner
membrane tend to associate with each other in ways that facilitate the
correct transfer of electrons between appropriate complexes.
THOUGHT PROBLEMS
14–24
Both H+ and Ca2+ are ions that move through the cytosol. Why is the
movement of H+ ions so much faster than that of Ca2+ ions? How do you
suppose the speed of these two ions would be afected by freezing the
solution? Would you expect them to move faster or slower? Explain your
answer.
14–25
Distinguish between a hydrogen atom, a proton, a hydride ion, and a
hydrogen molecule.
14–26
he half reactions for some of the carriers in the respiratory chain are
given in Table 14–1. From their E0ʹ values, what would you guess is their
order in the chain? What would you need to know before you were more
certain of their order?
14–27
he cytochrome c oxidase complex is strongly inhibited by cyanide,
which binds to the Fe3+ form of cytochrome a3. Cyanide kills at very low
concentration because of its efects on the cytochrome c oxidase complex. Carbon monoxide (CO) also inhibits the cytochrome c oxidase
287
288
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
TABLE 14–1 Standard redox potentials for electron carriers in the
respiratory chain (Problem 14–26).
Half reaction
E0ʹ (V)
ubiquinone + 2 H+ + 2 e– → ubiquinol
0.045
cytochrome b (Fe3+) + e– → cytochrome b (Fe2+)
0.077
cytochrome c1 (Fe3+) + e– → cytochrome c1 (Fe2+)
0.22
cytochrome c (Fe3+) + e– → cytochrome c (Fe2+)
0.25
cytochrome a (Fe3+) + e– → cytochrome a (Fe2+)
0.29
cytochrome a3 (Fe3+) + e– → cytochrome a3 (Fe2+)
0.55
complex by binding to cytochrome a3, but it binds to the Fe2+ form. CO
kills only at much higher doses than cyanide, not because of its binding
to the cytochrome c oxidase complex, but because it binds to the heme
group of hemoglobin, which also carries an Fe2+. In a sense, by mopping
up CO, hemoglobin protects cytochrome c oxidase from being inhibited
by CO at low concentrations. One treatment for cyanide poisoning—if
administered quickly enough—is to give sodium nitrite, which oxidizes
Fe2+ to Fe3+. How do you suppose sodium nitrite protects against the
efects of cyanide?
14–28
he two diferent difusible electron carriers, ubiquinone and cytochrome
c, shuttle electrons between the three protein complexes of the electrontransport chain. In principle, could the same difusible carrier be used
for both steps? If not, why not? If it could, what characteristics would it
need to possess and what would be the disadvantages of such a situation?
14–29
If you were to impose an artiicially large electrochemical gradient across
the mitochondrial inner membrane, would you expect electrons to low
up the respiratory chain, in the reverse of their normal direction? Why or
why not?
14–30
Some bacteria have become specialized to live in an alkaline environment at pH 10. hey maintain their internal environment at pH 7. Why
is it that they cannot exploit the pH diference across their membrane to
get ATP for free using a standard ATP synthase? Can you suggest an engineering modiication to ATP synthase that would allow it to generate ATP
from proton low in such an environment?
CALCULATIONS
14–31
One of the problems in understanding redox reactions is coming to grips
with the language. Consider the reduction of pyruvate by NADH:
pyruvate + NADH + H+
lactate + NAD+
In redox reactions, oxidation and reduction necessarily occur together;
however, it is convenient to list the two halves of a redox reaction separately. By convention, each half reaction is written as a reduction: oxidant
+ e– → reductant. For the reduction of pyruvate by NADH the half reactions are
pyruvate + 2 H+ + 2e– → lactate
E0ʹ = –0.19 V
NAD+ + H+ + 2e– → NADH
E0ʹ = –0.32 V
THE PROTON PUMPS OF THE ELECTRON-TRANSPORT CHAIN
where E0ʹ is the standard redox potential and refers to a reaction occurring under standard conditions (25°C or 298 K, all concentrations at 1
M, and pH 7). To obtain the overall equation for reduction of pyruvate
by NADH, it is necessary to reverse the NAD+/NADH half reaction and
change the sign of E0ʹ:
pyruvate + 2 H+ + 2e– → lactate
E0ʹ = –0.19 V
NADH → NAD+ + H+ + 2e–
E0ʹ = +0.32 V
Summing these two half reactions and their E0ʹ values gives the overall
equation and its ΔE0ʹ value:
pyruvate + NADH + H+
lactate + NAD+
ΔE0ʹ = +0.13 V
When a redox reaction takes place under nonstandard conditions, the
tendency to donate electrons (ΔE ) is equal to ΔE0ʹ modiied by a concentration term:
∆E = ∆E0′ –
2.3 RT
[lactate][NAD+]
log
nF
[pyruvate][NADH]
where R = 8.3 × 10–3 kJ/K mole, T = temperature in kelvins, n = the number of electrons transferred, and F = 96 kJ/V mole.
ΔG is related to ΔE by the equation
ΔG = –nF ΔE
Since the signs of ΔG and ΔE are opposite, a favorable redox reaction has
a positive ΔE and a negative ΔG.
A. Calculate ΔG for reduction of pyruvate to lactate at 37°C with all reactants
and products at a concentration of 1 M.
B. Calculate ΔG for the reaction at 37°C under conditions where the concentrations of pyruvate and lactate are equal and the concentrations of
NAD+ and NADH are equal.
C. What would the concentration term need to be for this reaction to have a
ΔG of zero at 37°C?
D. Under normal conditions in vascular smooth muscle (at 37°C), the concentration ratio of NAD+ to NADH is 1000, the concentration of lactate is
0.77 μmol/g, and the concentration of pyruvate is 0.15 μmol/g. What is
ΔG for reduction of pyruvate to lactate under these conditions?
14–32
hiobacillus ferrooxidans, a bacterium that lives on slag heaps at pH
2, is used by the mining industry to recover copper and uranium from
low-grade ore by an acid leaching process. he bacteria oxidize Fe2+ to
produce Fe3+, which in turn oxidizes (and solubilizes) these minor components of the ore. It is remarkable that the bacterium can live in such
an environment. It does so by exploiting the pH diference between the
environment and its cytoplasm (pH 6.5) to drive synthesis of ATP and
NADPH, which it can then use to ix CO2 and nitrogen. In order to keep
its cytoplasmic pH constant, T. ferrooxidans uses electrons from Fe2+ to
reduce O2 to water, thereby removing the protons:
4 Fe2+ + O2 + 4 H+ → 4 Fe3+ + 2 H2O
What are the energetics of these processes? Is the low of electrons from
Fe2+ to O2 energetically favorable? How diicult is it to reduce NADP+
using electrons from Fe2+? hese are key questions for understanding
how T. ferrooxidans manages to thrive in such an unlikely niche.
A. What are ΔE and ΔG for the reduction of O2 by Fe2+, assuming that the
reaction occurs under standard conditions? he half reactions are
Fe3+ + e– → Fe2+
E0ʹ = 0.77 V
O2 + 4 H+ + 4e– → 2 H2O
E0ʹ = 0.82 V
289
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
290
B. Write a balanced equation for the reduction of NADP+ + H+ by Fe2+. What
is ΔG for this reaction under standard conditions? he half reaction for
NADP+ is
NADP+ + H+ + 2e– → NADPH
E0ʹ = –0.32 V
What is ΔG for the reduction of NADP+ + H+ by Fe2+ if the concentrations
of Fe3+ and Fe2+ are equal, the concentration of NADPH is 10-fold greater
than that of NADP+, and the temperature is 310 K? (Note: adjusting the
number of atoms and electrons in order to balance the chemical equation does not afect E0ʹ or ΔE0ʹ values. It does, however, afect ΔE by its
inluence on the concentration term; each concentration term must be
raised to an exponent equal to the number of atoms or molecules used in
the balanced equation.)
14–33
What is the standard free-energy change (ΔG°) associated with transfer of electrons from NADH to O2, according to the balanced equation
below?
O2 + 2 NADH + 2 H+
2 H2O + 2 NAD+
he half reactions are
O2 + 4 H+ + 4e– → 2 H2O
E0ʹ = 0.82V
NAD+ + H+ + 2e– → NADH
E0ʹ = –0.32 V
A common way of writing this equation is
½ O2 + NADH + H+
H2O + NAD+
What is ΔG° for this equation? Do the two calculations give the same
answer? Explain why they do or don’t.
DATA HANDLING
14–34
In 1925, David Keilin used a simple spectroscope to observe the characteristic absorption bands of the cytochromes that make up the electrontransport chain in mitochondria. A spectroscope passes a very bright
light through the sample of interest and then through a prism to display
the spectrum from red to blue. If molecules in the sample absorb light of
particular wavelengths, dark bands will interrupt the colors of the rainbow. Keilin found that tissues from a wide variety of animals all showed
the pattern in Figure 14–2. (his pattern had actually been observed
several decades before by an Irish physician named MacMunn, but he
thought all the bands were due to a single pigment. His work was all but
forgotten by the 1920s.)
he diferent heat stabilities of the individual absorption bands and
their diferent intensities in diferent tissues led Keilin to conclude that
the absorption pattern was due to three components, which he labeled
cytochromes a, b, and c (Figure 14–2). His key discovery was that the
absorption bands disappeared when oxygen was introduced (Figure
14–3A) and then reappeared when the samples became anaerobic
cytochrome
absorption bands
c b
a
400
500
600
wavelength (nm)
700
Figure 14–2 Cytochrome absorption bands
(Problem 14–34).
THE PROTON PUMPS OF THE ELECTRON-TRANSPORT CHAIN
291
Figure 14–3 Cytochrome absorption
bands under a variety of experimental
conditions (Problem 14–34).
(A) AEROBIC
(B) ANAEROBIC
(C) AEROBIC PLUS CYANIDE
(D) AEROBIC PLUS URETHANE
(E) CYTOCHROME c PLUS OXYGEN
400
500
600
700
wavelength (nm)
14–35
If isolated mitochondria are incubated with a source of electrons such
as succinate, but without oxygen, electrons enter the respiratory chain,
reducing each of the electron carriers almost completely. When oxygen is
then introduced, the carriers become oxidized at diferent rates (Figure
14–4). How does this result allow you to order the electron carriers in the
respiratory chain? What is their order?
Inhibitors have provided extremely useful tools for analyzing mitochondrial function. Figure 14–5 shows three distinct patterns of oxygen electrode traces obtained using a variety of inhibitors. In all experiments,
mitochondria were added to a phosphate-bufered solution containing
succinate as the sole source of electrons for the respiratory chain. After
a short interval, ADP was added followed by an inhibitor, as indicated
in Figure 14–5. he rates of oxygen consumption at various times during
the experiment are shown by downward-sloping lines, with faster rates of
consumption shown by steeper lines.
A. Based on the descriptions of the inhibitors in Table 14–2, assign each
inhibitor to one of the oxygen traces in Figure 14–5. All these inhibitors
stop ATP synthesis.
B. Using the same experimental protocol indicated in Figure 14–5, sketch
14–36
reduced cytochrome (percent)
(Figure 14–3B). He later confessed, “his visual perception of an intracellular respiratory process was one of the most impressive spectacles I
have witnessed in the course of my work.”
Keilin subsequently discovered that cyanide prevented the bands
from disappearing when oxygen was introduced (Figure 14–3C). When
urethane (an inhibitor of electron transport that is no longer used) was
added, bands a and c disappeared in the presence of oxygen, but band
b remained (Figure 14–3D). Finally, using cytochrome c extracted from
dried yeast, he showed that the band due to cytochrome c remained
when oxygen was present (Figure 14–3E).
A. Is it the reduced (electron-rich) or the oxidized (electron-poor) forms of
the cytochromes that give rise to the bands that Keilin observed?
B. From Keilin’s observations, deduce the order in which the three
cytochromes carry electrons from intracellular substrates to oxygen.
C. One of Keilin’s early observations was that the presence of excess glucose prevented the disappearance of the absorption bands when oxygen
was added. How do you think that rapid glucose oxidation to CO2 might
explain this observation?
100
add
O2
b
c1
50
c
add
succinate
(a + a3)
0
time
Figure 14–4 Rapid spectrophotometric
analysis of the rates of oxidation of
electron carriers in the respiratory chain
(Problem 14–35). Cytochromes a and a3
cannot be distinguished and thus are
listed as cytochrome (a + a3).
292
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
TABLE 14–2 Effects of a variety of inhibitors of mitochondrial function
(Problem 14–36).
Inhibitor
Function
1. FCCP
Makes membranes permeable to protons
2. Malonate
Prevents oxidation of succinate
3. Cyanide
Inhibits the cytochrome c oxidase complex
4. Atractyloside
Inhibits the ADP/ATP transporter
5. Oligomycin
Inhibits ATP synthase
6. Butylmalonate
Blocks mitochondrial uptake of succinate
(A)
mitochondria
ADP
inhibitor
(B)
mitochondria
ADP
inhibitor
(C)
mitochondria
ADP
the oxygen traces that you would expect for the sequential addition of the
pairs of inhibitors in the list below.
1. FCCP followed by cyanide
2. FCCP followed by oligomycin
3. Oligomycin followed by FCCP
inhibitor
MEDICAL LINKS
14–37
Normally, the low of electrons to O2 is tightly linked to the production
of ATP via the electrochemical gradient. If ATP synthase is inhibited, for
example, electrons do not low down the electron-transport chain and
respiration ceases. Since the 1940s, several substances—such as 2,4-dinitrophenol—have been known to uncouple electron low from ATP synthesis. Dinitrophenol was once prescribed as a diet drug to aid in weight
loss. How would an uncoupler of oxidative phosphorylation promote
weight loss? Why do you suppose dinitrophenol is no longer prescribed?
ATP PRODUCTION IN MITOCHONDRIA
TERM TO LEARN
ATP synthase
DEFINITIONS
Match the deinition below with its term from the list above.
14–38
Enzyme in the inner membrane of a mitochondrion that catalyzes the
formation of ATP from ADP and inorganic phosphate.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
14–39
An average person contains 50 kg of ATP, which is required to meet their
daily energy needs.
14–40
he number of c subunits in the rotor ring of ATP synthase deines how
many protons need to pass through the turbine to make each molecule of
ATP.
THOUGHT PROBLEMS
14–41
he respiratory chain is relatively inaccessible to experimental manipulation in intact mitochondria. After disrupting mitochondria with ultrasound, however, it is possible to isolate functional submitochondrial particles, which consist of broken cristae that have resealed inside-out into
small closed vesicles. In these vesicles, the components that originally
Figure 14–5 oxygen traces showing three
patterns of inhibitor effects on oxygen
consumption by mitochondria (Problem
14–36).
ATP PRODUCTION IN MITOCHONDRIA
293
faced the matrix are now exposed to the surrounding medium. How do
you suppose such an arrangement might aid in the study of electron
transport and ATP synthesis?
14–42
As electrons move down the respiratory chain, protons are pumped
across the inner membrane. Are those protons conined to the intermembrane space? Why or why not?
You have reconstituted into the membranes of the same vesicles puriied
bacteriorhodopsin, which is a light-driven H+ pump from a photosynthetic bacterium, and puriied ATP synthase from ox heart mitochondria.
Assume that all molecules of bacteriorhodopsin and ATP synthase are
oriented as shown in Figure 14–6, so that protons are pumped into the
vesicle and ATP synthesis occurs on the outer surface.
A. If you add ADP and phosphate to the external medium and shine light
into the suspension of vesicles, would you expect ATP to be generated?
Why or why not?
B. If you prepared the vesicles without being careful to remove all the detergent, which makes the bilayer leaky to protons, would you expect ATP to
be synthesized?
C. If the ATP synthase molecules were randomly oriented so that about
half faced the outside of the vesicle and half faced the inside, would you
expect ATP to be synthesized? If the bacteriorhodopsin molecules were
randomly oriented, would you expect ATP to be synthesized? Explain
your answers.
D. You tell a friend over dinner about your new experiments. He questions
the validity of an approach that utilizes components from so widely
divergent, unrelated organisms. As he so succinctly puts it, “Why would
anybody want to mix vanilla pudding with brake luid?” Defend your
approach against his criticism.
+ detergent
bacteriorhodopsin
ADD PHOSPHOLIPIDS,
14–43
14–44
An elongated arm—the stator—links the catalytic head group (the α3β3
complex) of the ATP synthase to the membrane-embedded rotor component. Attached to the rotor is a stalk (the axle-like γ subunit) that turns
inside the head group to force the conformational changes that lead to
ATP synthesis. If the stator were missing, would ATP be synthesized in
response to the proton low? Why or why not?
CALCULATIONS
14–45
In actively respiring liver mitochondria, the pH in the matrix is about half
a pH unit higher than it is in the cytosol. Assuming that the cytosol is at
pH 7 and the matrix is a sphere with a diameter of 1 μm [V = (4/3) πr3],
calculate the total number of protons in the matrix of a respiring liver
mitochondrion. If the matrix began at pH 7 (equal to that in the cytosol),
how many protons would have to be pumped out to establish a matrix pH
of 7.5 (a diference of 0.5 pH unit)?
14–46
he relationship of free-energy change (ΔG) to the concentrations of
reactants and products is important because it predicts the direction
of spontaneous chemical reactions. Familiarity with this relationship is
essential for understanding energy conversions in cells. Consider, for
example, the hydrolysis of ATP to ADP and inorganic phosphate (Pi):
ATP + H2O → ADP + Pi
he free-energy change due to ATP hydrolysis is
[ADP][Pi]
[ATP]
[ADP][Pi]
= ∆G °+ 2.3 RT log
[ATP]
∆G = ∆G °+ RT ln
ATP synthase
REMOVE DETERGENT
LIGHT
H+
sealed
vesicle
Figure 14–6 Reconstitution of
bacteriorhodopsin and ATP synthase
into lipid vesicles (Problem 14–43).
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
294
where the concentrations are expressed as molarities (by convention,
the concentration of water is not included in the expression). R is the gas
constant (8.3 × 10–3 kJ/K mole), T is temperature (assume 37°C, which
is 310 K), and ΔG° is the standard free-energy change (–30.5 kJ/mole for
ATP hydrolysis to ADP and Pi).
A. Calculate ΔG for ATP hydrolysis when the concentrations of ATP, ADP,
and Pi are all equal to 1 M. What is ΔG when the concentrations of ATP,
ADP, and Pi are all equal to 1 mM?
B. In a resting muscle, the concentrations of ATP, ADP, and Pi are approximately 5 mM, 1 mM, and 10 mM, respectively. What is ΔG for ATP hydrolysis in resting muscle?
C. What will ΔG equal when the hydrolysis reaction reaches equilibrium? At
[Pi] = 10 mM, what will be the ratio of [ATP] to [ADP] at equilibrium?
D. Show that, at constant [Pi], ΔG decreases by 5.9 kJ/mole for every 10-fold
increase in the ratio of [ATP] to [ADP], regardless of the value of ΔG°. (For
example, ΔG decreases by 11.8 kJ/mole for a 100-fold increase, by 17.7
kJ/mole for a 1000-fold increase, and so on.)
DATA HANDLING
ATP synthase is the world’s smallest rotary motor. Passage of H+ ions
through the membrane-embedded portion of ATP synthase (the Fo component) causes rotation of the single, central, axle-like γ subunit inside
the head group. he tripartite head is composed of the three αβ dimers,
the β subunit of which is responsible for synthesis of ATP. he rotation
of the γ subunit induces conformational changes in the αβ dimers that
allow ADP and Pi to be converted into ATP. A variety of indirect evidence
had suggested rotary catalysis by ATP synthase, but seeing is believing.
To demonstrate rotary motion, a modiied form of the α3β3γ complex
was used. he β subunits were modiied so they could be irmly anchored
to a solid support and the γ subunit was modiied (on the end that normally inserts into the Fo component in the inner membrane) so that a
luorescently tagged, readily visible ilament of actin could be attached
(Figure 14–7A). his arrangement allows rotations of the γ subunit to
be visualized as revolutions of the long actin ilament. In these experiments, ATP synthase was studied in the reverse of its normal mechanism
by allowing it to hydrolyze ATP. At low ATP concentrations, the actin ilament was observed to revolve in steps of 120° and then pause for variable
lengths of time, as shown in Figure 14–7B.
A. Why does the actin ilament revolve in steps with pauses in between?
What does this rotation correspond to in terms of the structure of the
α3β3γ complex?
B. In its normal mode of operation inside the cell, how many ATP molecules
do you suppose would be synthesized for each complete 360° rotation of
the γ subunit? Explain your answer.
14–47
(A)
(B)
direction of rotation
5
inner membrane
matrix
β
β
α
revolutions
4
actin filament
γ
3
2
1
0
0
20
40
60
80
time (seconds)
100
Figure 14–7 Experimental set-up for
observing rotation of the γ subunit
of ATP synthase (Problem 14–47).
(A) The immobilized α3β3γ complex.
The β subunits are anchored to
a solid support and a luorescent
actin ilament is attached to the γ
subunit. (B) stepwise revolution of
the actin ilament. The indicated
trace is a typical example from one
experiment. The inset shows the
positions in the revolution at which
the actin ilament pauses.
ATP PRODUCTION IN MITOCHONDRIA
295
(B)
pt
y
i
P
AT
P
P
ATP
14–48
he three αβ dimers in each ATP synthase normally exist in three different conformations: one empty, one with ADP and Pi bound, and one
with ATP bound. he conformational changes are driven sequentially by
rotation of the γ subunit, which in turn is driven by the low of protons
through the ATP synthase. As viewed from the inner membrane, looking at the underside of the α3β3 complex, these sites could be arranged
in either of two ways around the central γ subunit (Figure 14–8). he
sequential, linked conformational changes driven by proton low are
also shown for the two arrangements in the igure. In Problem 14–47, the
revolutions of the attached actin ilaments during ATP hydrolysis were
shown to be counterclockwise when viewed from the same perspective
(see Figure 14–7). Which of the two arrangements of conformations of αβ
dimers shown in Figure 14–8 is correct? Explain your answer.
14–49
A manuscript has been submitted to a prestigious scientiic journal. In
it the authors describe an experiment using an immobilized α3β3γ complex with an attached actin ilament like that shown in Figure 14–7A. he
authors show that they can mechanically rotate the γ subunit by applying
force to the actin ilament. Moreover, in the presence of ADP and phosphate, each 120° clockwise rotation of the γ subunit is accompanied by
the synthesis of one molecule of ATP. Is this result at all reasonable? What
would such an observation imply about the mechanism of ATP synthase?
Should this manuscript be considered for publication in one of the best
journals?
14–50
he ADP/ATP transporter in the mitochondrial inner membrane can
exchange ATP for ATP, ADP for ADP, and ATP for ADP. Even though mitochondria can transport both ADP and ATP, there is a strong bias in favor
of exchange of external ADP for internal ATP in actively respiring mitochondria. You suspect that this bias is due to the conversion of ADP into
ATP inside the mitochondrion. ATP synthesis would continually reduce
the internal concentration of ADP and thereby create a favorable concentration gradient for import of ADP. he same process would increase the
internal concentration of ATP, thereby creating favorable conditions for
export of ATP.
To test your hypothesis, you conduct experiments on isolated mitochondria. In the absence of substrate (when the mitochondria are not
respiring and the membrane is uncharged), you ind that ADP and ATP
are taken up at the same rate. When you add substrate, the mitochondria begin to respire, and ADP enters mitochondria at a much faster rate
than ATP. As you expected, when you add dinitrophenol, which collapses
the pH gradient, along with the substrate, ADP and ATP again enter at
Pi
ATP
ADP + Pi
P
pt
y
3H+
AD
em
ADP Pi
y
pt
Pi
ADP + Pi
ATP
3H+
ATP
em
P
3H+
ADP + Pi
AD
y
pt
ADP Pi
3H+
ATP
ADP + Pi
AD
empty
3H+
em
AT
P
ATP
em
AD
P
P
AT
ATP
ADP + Pi
P
AT
P
i
(A)
3H+
ADP + Pi
empty
ATP
Figure 14–8 The two possible
arrangements of conformations of the
three αβ dimers in ATP synthase, along
with the linked conformational changes
driven by proton low (Problem 14–48).
One αβ dimer is colored red to emphasize
that its position remains ixed as it
changes conformation in response to
proton low. The perspective illustrated is
from the inner membrane, looking at the
underside of the tripartite α3β3 complex.
296
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
Experiment
1
Substrate
Absent
Inhibitor
None
Relative rates
of entry
ADP = ATP
2
Present
None
ADP > ATP
3
Present
Dinitrophenol
ADP = ATP
4
Present
Oligomycin
ADP > ATP
O
–
O
P
O
O
O–
P
photochemical reaction center
photosynthetic electron transfer
photosystem
stroma
thylakoid membrane
DEFINITIONS
Match each deinition below with its term from the list above.
14–51
Light-driven reactions in photosynthesis in which electrons move along
the electron-transport chain in the thylakoid membrane.
14–52
Part of a photosystem that captures light energy and channels it into the
photochemical reaction center.
14–53
Process by which green plants incorporate carbon atoms from atmospheric carbon dioxide into sugars.
14–54
he part of a photosystem that converts light energy into chemical energy.
14–55
Organelle in green algae and plants that contains chlorophyll and carries
out photosynthesis.
14–56
Light-absorbing green pigment that plays a central role in photosynthesis.
14–57
he large space that surrounds the inner chloroplast membrane.
P
N
N
O
CH2
O–
O
H
H
H
H
OH OH
NH2
ADP
N
N
O
P
O–
TERMS TO LEARN
antenna complex
carbon fixation
carbon-fixation reactions
charge separation
chlorophyll
chloroplast
O
O
O–
–
O
CHLOROPLASTS AND PHOTOSYNTHESIS
N
N
In all cases, the initial rates of entry of ATP and ADP were measured.
the same rate. When you add an inhibitor of ATP synthase (oligomycin)
along with the substrate, ADP is taken up much faster than ATP. Your
results are summarized in Table 14–3. You are puzzled by the results
with oligomycin, since your hypothesis predicted that the rates of uptake
would be equal.
When you show the results to your advisor, she compliments you
on your ine experiments and agrees that they disprove the hypothesis.
She suggests that you examine the structures of ATP and ADP (Figure
14–9) if you wish to understand the behavior of the ADP/ATP transporter.
What is the correct explanation for the biased exchange by the ADP/ATP
transporter under some of the experimental conditions and an unbiased
exchange under others?
NH2
ATP
TABLE 14–3 Entry of ADP and ATP into isolated mitochondria (Problem
14–50).
O
O
P
N
N
O
CH2
O–
H
H
O
H
H
OH OH
Figure 14–9 Structures of ATP and ADP
(Problem 14–50).
CHLOROPLASTS AND PHOTOSYNTHESIS
297
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
14–58
In a general way, one might view the chloroplast as a greatly enlarged
mitochondrion in which the cristae have been pinched of to form a
series of interconnected submitochondrial particles in the matrix space.
14–59
When a molecule of chlorophyll in an antenna complex absorbs a photon, the excited electron is rapidly transferred from one chlorophyll molecule to another until it reaches the photochemical reaction center.
THOUGHT PROBLEMS
14–60
Both mitochondria and chloroplasts use electron transport to pump protons, creating an electrochemical proton gradient, which drives ATP synthesis. Are protons pumped across the same (analogous) membranes in
the two organelles? Is ATP synthesized in the analogous compartments?
Explain your answers.
14–61
A suspension of the green alga Chlamydomonas is actively carrying out
photosynthesis in the presence of light and CO2. If you turned of the
light, how would you expect the amounts of ribulose 1,5-bisphosphate
and 3-phosphoglycerate to change over the next minute? How about if
you left the light on but removed the CO2?
14–62
Why are plants green?
14–63
Treatment of chloroplasts with the herbicide DCMU stops O2 evolution
and photophosphorylation. If an artiicial electron acceptor is added that
accepts electrons from plastoquinone (Q), oxygen evolution is restored
but not photophosphorylation. Propose a site at which DCMU acts in the
low of electrons through photosystems II and I (Figure 14–10). Explain
your reasoning. Why is DCMU an herbicide?
14–64
In chloroplasts, protons are pumped out of the stroma across the thylakoid membrane, whereas in mitochondria, they are pumped out of the
matrix across the crista membrane. Explain how this arrangement allows
chloroplasts to generate a larger proton gradient across the thylakoid
membrane than mitochondria can generate across the inner membrane.
14–65
Unlike mitochondria, chloroplasts do not have a transporter that allows
them to export ATP to the cytosol. How, then, does the rest of the cell get
the ATP it needs to survive?
CALCULATIONS
14–66
How much energy is available in visible light? How much energy does
sunlight deliver to Earth? How eicient are plants at converting light
energy into chemical energy? he answers to these questions provide an
important backdrop to the subject of photosynthesis.
Each quantum or photon of light has energy hv, where h is Planck’s
constant (6.6 × 10–37 kJ sec/photon) and v is the frequency in sec–1.
he frequency of light is equal to c/λ, where c is the speed of light
(3.0 × 1017 nm/sec) and λ is the wavelength in nm. hus, the energy (E)
of a photon is
E = hv = hc/λ
A. Calculate the energy of a mole of photons (6 × 1023 photons/mole) at
400 nm (violet light), at 680 nm (red light), and at 800 nm (near-infrared
light).
photosystem I
photosystem II
Q
light
separates
charges
H 2O
NADPH
light
separates
charges
cytochromes
pC
+
+
electron flow
Figure 14–10 Flow of electrons
through photosystems II and I during
photosynthesis in chloroplasts (Problem
14–63). Electrons from photosystem II
low to plastoquinone (Q), then to the
cytochrome b6-f complex (cytochromes),
and then to plastocyanin (pC), after which
they enter photosystem I. The protons
pumped by the cytochrome b6-f complex
generate an electrochemical gradient,
which is used to drive aTp synthesis.
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
298
B. Bright sunlight strikes Earth at the rate of about 1.3 kJ/sec per square
meter. Assuming for the sake of calculation that sunlight consists of monochromatic light of wavelength 680 nm, how many seconds would it take
for a mole of photons to strike a square meter?
C. Assuming that it takes eight photons to ix one molecule of CO2 as carbohydrate under optimal conditions (8–10 photons is the currently
accepted value), calculate how long it would take a tomato plant with a
leaf area of 1 square meter to make a mole of glucose from CO2. Assume
that photons strike the leaf at the rate calculated above and, furthermore,
that all the photons are absorbed and used to ix CO2.
D. If it takes 468 kJ/mole to ix a mole of CO2 into carbohydrate, what is the
eiciency of conversion of light energy into chemical energy after photon capture? Assume again that eight photons of red light (680 nm) are
required to ix one molecule of CO2.
14–67
14–68
What fraction of the free energy of light at 700 nm is captured when a
chlorophyll molecule (P700) at the photochemical reaction center in
photosystem I absorbs a photon? he equation for calculating the free
energy available in one photon of light is given in Problem 14–66. If one
assumes standard conditions, the captured free energy (ΔG = –nFΔE0ʹ)
can be calculated from the standard redox potential for P700* (excited)
→ P700 (ground state), which can be gotten from the half reactions:
P700+ + e– → P700
E0ʹ = 0.4 V
P700+ + e– → P700*
E0ʹ = –1.2 V
he balanced equation for production of NADPH by the Z scheme of
photophosphorylation is
2 H2O + 2 NADP+ → 2 NADPH + 2 H+ + O2
How many photons must be absorbed to generate two NADPH and a
molecule of O2? (Assume one photon excites one electron.)
14–69
T. ferrooxidans, the slag-heap bacterium that lives at pH 2, ixes CO2 like photosynthetic organisms but uses the abundant Fe2+ in its environment as a
source of electrons instead of H2O. T. ferrooxidans oxidizes Fe2+ to Fe3+ to
reduce NADP+ to NADPH, a very unfavorable reaction with a ΔE of about
–1.1 V. It does so by coupling production of NADPH to the energy of the natural proton gradient across its membrane, which has a free-energy change
(ΔG) of –26.8 kJ/mole H+. What is the smallest number of protons to the
nearest integer that would be required to drive the reduction of NADP+ by
Fe2+? How do you suppose proton low is mechanistically coupled to the
reduction of NADP+?
DATA HANDLING
14–70
Careful experiments comparing absorption and action spectra of plants
ultimately led to the notion that two photosystems cooperate in chloroplasts. he absorption spectrum is the amount of light captured by photosynthetic pigments at diferent wavelengths. he action spectrum is the
rate of photosynthesis (for example, O2 evolution or CO2 ixation) resulting from the capture of photons.
T.W. Engelmann, who used simple equipment and an ingenious
experimental design, probably made the irst measurement of an action
spectrum in 1882. He placed a ilamentous green alga into a test tube
along with a suspension of oxygen-seeking bacteria. He allowed the bacteria to use up the available oxygen and then illuminated the alga with
light that had been passed through a prism to form a spectrum. After a
short time he observed the results shown in Figure 14–11. Sketch the
CHLOROPLASTS AND PHOTOSYNTHESIS
299
Figure 14–11 Experiment to measure the
action spectrum of a ilamentous green
alga (Problem 14–70). Bacteria, which are
indicated by the brown rectangles, were
distributed evenly throughout the test
tube at the beginning of the experiment.
400
ultraviolet
500
blue
green
600
yellow
700
red
wavelength (nm)
infrared
action spectrum (O2 evolved at diferent wavelengths of light) for this
alga and explain how this experiment works.
he most compelling early evidence for the Z scheme of photosynthesis
came from measuring the oxidation states of the cytochromes in algae
under diferent regimes of illumination (Figure 14–12). Illumination
with light at 680 nm caused oxidation of cytochromes (indicated by the
upward trace in Figure 14–12A). Additional illumination with light at
562 nm caused reduction of the cytochromes (indicated by the downward trace in Figure 14–12A). When the lights were then turned of, both
efects were reversed (Figure 14–12A). In the presence of the herbicide
DCMU (see Problem 14–63), no reduction with 562-nm light occurred
(Figure 14–12B).
A. In these algae, which wavelength stimulates photosystem I and which
stimulates photosystem II?
B. How do these results support the Z scheme for photosynthesis; that is,
how do they support the idea that there are two photosystems that are
linked by cytochromes?
C. On which side of the cytochromes does DCMU block electron transport—on the side nearer photosystem I or the side nearer photosystem
II?
14–71
Photosystem II accepts electrons from water, generating O2, and donates
them via the electron-transport chain to photosystem I. Each photon
absorbed by photosystem II transfers only a single electron, and yet four
(A)
680
on
562
on
562
off
680
off
(B)
680
on
562
on
562
off
680
off
absorbance at 420 nm
absorbance at 420 nm
14–72
more oxidized
DCMU
more reduced
0
5
10
time (seconds)
15
20
Figure 14–12 Oxidation state of
cytochromes after illumination of algae
with light of different wavelengths
(Problem 14–71). (a) In the absence of
DCMU. (B) In the presence of DCMU. an
upward trace indicates oxidation of the
cytochromes; a downward trace indicates
reduction of the cytochromes.
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
Figure 14–13 oxygen evolution by spinach chloroplasts in response to
saturating lashes of light (Problem 14–72). The chloroplasts were placed
in the dark for 40 minutes prior to the experiment to allow them to come
to the same “ground” state. Oxygen production is expressed in arbitrary
units.
electrons must be removed from water to generate a molecule of O2.
hus, four photons are required to produce a molecule of O2:
80
oxygen produced
300
60
40
20
0
2 H2O + 4 hv → 4e– + 4 H+ + O2
0
How do four photons cooperate in the production of O2? Is it necessary
that four photons arrive at a single reaction center simultaneously? Can
four activated reaction centers cooperate to produce a molecule of O2?
Or is there some sort of “gear wheel” that collects the four electrons from
H2O and transfers them one at a time to a reaction center?
To investigate this problem, you expose dark-adapted spinach chloroplasts to a series of brief saturating lashes of light (2 μsec) separated
by short periods of darkness (0.3 sec) and measure the production of O2
that results from each lash. Under this lighting regime, most photosystems capture a photon during each lash. As shown in Figure 14–13, O2
is produced with a distinct periodicity: the irst burst of O2 occurs on the
third lash, and subsequent peaks occur every fourth lash thereafter. If
you irst inhibit 97% of the photosystem II reaction centers with DCMU
and then repeat the experiment, you observe the same periodicity of O2
production, but the peaks are only 3% of the uninhibited values.
A. How do these results distinguish among the three possibilities posed at
the outset (simultaneous action, cooperation among reaction centers,
and a gear wheel)?
B. Why do you think it is that the irst burst of O2 occurs after the third lash,
whereas additional peaks occur at four-lash intervals? (Consider what
this observation implies about the dark-adapted state of the chloroplasts.)
C. Can you suggest a reason why the periodicity in O2 production becomes
less pronounced with increasing lash number?
2
4
6
8 10 12 14
flash number
In an insightful experiment performed in the 1960s, chloroplasts were
irst soaked in an acidic solution at pH 4, so that the stroma and thylakoid
space became acidiied (Figure 14–14). hey were then transferred to a
basic solution (pH 8). his rapidly increased the pH of the stroma to 8,
while the thylakoid space temporarily remained at pH 4. A burst of ATP
synthesis was observed, and the pH diference between the thylakoid
space and the stroma quickly disappeared.
A. Explain why these conditions lead to ATP synthesis.
B. Is light needed for the experiment to work? Why or why not?
C. What would happen if the solutions were switched so that the irst incubation was in the pH 8 solution and the second one was in the pH 4 solution? Explain your answer.
D. Does this experiment support the chemiosmotic model, or raise questions about it?
14–73
INCUBATE
CHLOROPLAST
FOR SEVERAL
HOURS
pH 4
CHANGE
EXTERNAL pH
AND ADD ADP
AND Pi
pH 4
pH 8
Figure 14–14 soaking chloroplasts in
acidic and basic solutions (Problem
14–73). Pink areas are at ph 4.
THE GENETIC SYSTEMS OF MITOCHONDRIA AND CHLOROPLASTS
THE GENETIC SYSTEMS OF MITOCHONDRIA AND
CHLOROPLASTS
TERM TO LEARN
maternal inheritance
DEFINITIONS
Match the deinition below with its term from the list above.
14–74
Pattern of mitochondrial inheritance in higher animals that arises
because the egg cells always contribute much more cytoplasm to the
zygote than does the sperm.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
14–75
he mitochondrial genetic code difers slightly from the nuclear code,
but it is identical in mitochondria from all species that have been examined.
14–76
he presence of introns in organellar genes is not surprising since similar
introns have been found in related genes from bacteria whose ancestors
are thought to have given rise to mitochondria and chloroplasts.
14–77
Mutations that are inherited according to Mendelian rules afect nuclear
genes; mutations whose inheritance violates Mendelian rules are likely
to afect organelle genes.
THOUGHT PROBLEMS
14–78
You have discovered a remarkable, one-celled protozoan that lives in an
anaerobic environment. It has caught your attention because it has absolutely no mitochondria, an exceedingly rare condition among eukaryotes. If you could show that this organism derives from an ancient lineage that split of from the rest of eukaryotes before mitochondria were
acquired, it would truly be a momentous discovery. You sequence the
organism’s genome so you can make detailed comparisons. It is clear
from sequence comparisons that your organism does indeed derive from
an ancient lineage. But here and there, scattered around the genome, are
bits of DNA that in aggregate resemble the bacterial genome from which
mitochondria evolved. Propose a plausible evolutionary history for your
organism.
14–79
At the cellular level, evolutionary theories are particularly diicult to
test since fossil evidence is lacking. he possible evolutionary origins
of mitochondria and chloroplasts must be sought in living organisms.
Fortunately, living forms resembling the ancestral types thought to
have established an endosymbiotic relationship that led to the origin
of mitochondria and chloroplasts can be found today. For example,
the plasma membrane of the free-living aerobic bacterium Paracoccus
denitriicans contains a respiratory chain that is nearly identical to the
respiratory chain of mammalian mitochondria—both in the types of respiratory components present and in its sensitivity to respiratory inhibitors. Indeed, no signiicant feature of the mammalian respiratory chain
is absent from Paracoccus. Paracoccus efectively assembles in a single
organism all those features of the mitochondrial inner membrane that
are otherwise distributed at random among other aerobic bacteria.
301
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
302
(A)
(B)
(C)
Figure 14–15 A variegated leaf of Aucuba japonica with green and
yellow patches (Problem 14–80).
Imagine that you are a protoeukaryotic cell looking out for your evolutionary future. You have been observing proto-Paracoccus and are
amazed at its incredibly eicient use of oxygen in generating ATP. With
such a source of energy, your horizons would be unlimited. You plot to
hijack a proto-Paracoccus and make it work for you and your descendants. You plan to take it into your cytoplasm, feed it any nutrients it
needs, and harvest the ATP. Accordingly, one dark night, you trap a wandering proto-Paracoccus, surround it with your plasma membrane, and
imprison it in a new cytoplasmic compartment. To your relief, the protoParacoccus seems to enjoy its new environment. After a day of waiting,
however, you feel as sluggish as ever. What has gone wrong with your
scheme?
14–80
Examine the variegated leaf shown in Figure 14–15. Yellow patches
surrounded by green are common, but there are no green patches surrounded by yellow. Propose an explanation for this phenomenon.
14–81
he pedigrees in Figure 14–16 show one example each of the following
types of mutation: mitochondrial mutation, autosomal recessive mutation, autosomal dominant mutation, and X-linked recessive mutation. In
each family, the parents have had nine children. Assign each pedigree to
one type of mutation. Explain the basis for your assignments.
(D)
Figure 14–16 Hypothetical pedigrees
representing four patterns of inheritance
(Problem 14–81). Males are shown as
squares; females as circles. Affected
individuals are shown as red symbols;
unaffected individuals are shown as white
symbols.
DATA HANDLING
It is well accepted that transfer of DNA from organellar genomes to
nuclear genomes is common during evolution. Do transfers between
organellar genomes also occur? One experiment to search for genetic
transfers between organellar genomes used a deined restriction fragment from spinach chloroplasts, which carried information for the gene
for the large subunit of ribulose bisphosphate carboxylase. his gene has
no known mitochondrial counterpart. hus, if a portion of the chloroplast DNA in the restriction fragment were transferred to the mitochondrial genome, it would show up as a hybridizing band at a novel position.
Mitochondrial and chloroplast DNAs were prepared from zucchini, corn,
spinach, and pea. All these DNAs were digested with the same restriction
nuclease, and the resulting fragments were separated by electrophoresis.
he fragments were then transferred to a ilter and hybridized to a radioactive preparation of the spinach fragment. A schematic representation
of the autoradiograph is shown in Figure 14–17.
A. It is very diicult to prepare mitochondrial DNA that is not contaminated
to some extent with chloroplast DNA. How do these experiments control
for contamination of the mitochondrial DNA preparation by chloroplast
DNA?
B. Which of these plant mitochondrial DNAs appear to have acquired chloroplast DNA?
m
c
m
sp
c
m
c
a
pe
in
ac
h
rn
co
zu
cc
hi
ni
14–82
m
c
origin
Figure 14–17 Patterns of hybridization
of a probe from spinach chloroplast
DNA to mitochondrial and chloroplast
DNAs from zucchini, corn, spinach, and
pea (Problem 14–82). lanes labeled m
contain mitochondrial DNA; lanes labeled
c contain chloroplast DNA. Restriction
fragments to which the probe hybridized
are shown as dark bands.
THE GENETIC SYSTEMS OF MITOCHONDRIA AND CHLOROPLASTS
303
tRNA genes
mitochondrial DNA
F
V
L
IM W
D K
13
GR
16
12S 16S
ribosomal
RNA
HSL
T
7
mRNAs
Figure 14–18 Transcription map of human
mitochondrial DNA (Problem 14–83).
Individual tRNA genes are indicated by
red circles; the amino acids they carry are
shown in the one-letter code. The three
mRNAs whose detailed sequences are
shown in Figure 14–19 are indicated by
number.
he majority of mRNAs, tRNAs, and rRNAs in human mitochondria are
transcribed from one strand of the genome. hese RNAs are all present
initially on one very long transcript, which is 93% of the length of the DNA
strand. During mitochondrial protein synthesis, these RNAs function as
separate, independent species of RNA. he relationship of the individual RNAs to the primary transcript and many of the special features of
the mitochondrial genetic system have been revealed by comparing the
sequences of the RNAs with the nucleotide sequence of the genome. An
overview of the transcription map is shown in Figure 14–18.
hree segments of the nucleotide sequence of the human mitochondrial genome are shown in Figure 14–19 along with the three mRNAs
that are generated from those regions. he nucleotides that encode tRNA
species are highlighted; the amino acids encoded by the mRNAs are indicated below the center base of the codon.
A. In terms of codon usage and mRNA structure, in what two ways does initiation of protein synthesis in mitochondria difer from initiation in the
cytoplasm?
B. In what two ways are the termination codons for protein synthesis in
mitochondria unusual? (he termination codons are shown in Figure
14–19 as asterisks.)
C. Does the arrangement of tRNA and mRNA sequences in the genome
suggest a possible mechanism for processing the primary transcript into
individual RNA species?
14–83
14–84
A friend of yours has been studying a pair of mutants in the fungus Neurospora, which she has whimsically named poky and puny. Both mutants
grow at about the same rate, but much more slowly than wild type. Your
friend has been unable to ind any supplement that improves their
growth rates. Her biochemical analysis shows that each mutant displays
a diferent abnormal pattern of cytochrome absorption. To characterize
the mutants genetically, she crossed them to wild type and to each other
and tested the growth rates of the progeny. She has come to you because
she is puzzled by the results.
She explains that haploid nuclei from the two parents fuse during a
Neurospora mating and then divide meiotically to produce four haploid
spores, which can be readily tested for their growth rates. he parents
tRNAI
tRNAL
TTCTTAACAACATACCCAT.........CTCAAACCTAAGAAATATG
ACAUACCCAU.........CUCAAACCUAAAAAAAAAA
M P
E T *
tRNAD
mRNA 13
protein
tRNAK
TATATCTTAATGGCACATG.........CTCTAGAGCCCACTGTAAA
AUGGCACAUG.........CUCUAGAGCCAAAAAAAAA
M
tRNAR
DNA
A
H
S
DNA
mRNA 16
protein
*
tRNAH
ATTTACCAAATGCCCCTCA.........TTTTCCTCTTGTAAATATA
AUGCCCCUCA.........UUUUCCUCUUAAAAAAAAA
M P L
F S S *
DNA
mRNA 7
protein
Figure 14–19 Arrangements of tRNA
and mRNA sequences at three places
on the human mitochondrial genome
(Problem 14–83). Highlighted sequences
indicate tRNA genes. The sequences
of the mRNAs are shown in blue below
the corresponding genes. The middle
portions of the mRNAs and their genes
are indicated by dots. The 5ʹ ends of the
sequences are shown at the left. The
5ʹ ends of the mRNAs are unmodiied
and the 3ʹ ends have poly-a tails. The
encoded protein sequences are indicated
in green below the mrnas, with the letter
for the amino acid immediately under
the center nucleotide of the codon. an
asterisk (*) indicates a termination codon.
Chapter 14: Energy Conversion: Mitochondria and Chloroplasts
304
TABLE 14–4 Genetic analysis of Neurospora mutants (Problem 14–84).
Spore counts
Protoperithecial
parent
Fertilizing
parent
Fast growth
Slow growth
1
poky
wild
0
1749
2
wild
poky
1334
0
3
puny
wild
850
799
4
wild
puny
793
801
5
poky
puny
0
1831
6
puny
poky
754
710
7
wild
wild
1515
0
8
poky
poky
0
1389
9
puny
puny
0
1588
Cross
contribute unequally to the diploid: one parent (the protoperithecial
parent) donates a nucleus and the cytoplasm; the other (the fertilizing
parent) contributes little more than a nucleus—much like egg and sperm
in higher organisms. As shown in Table 14–4, the “order” of the crosses
sometimes makes a diference: this is a result she has not seen before.
Can you help your friend understand these results?
MCAT STYLE
Passage 1 (Questions 14–85 to 14–87)
Scientists discovered the mechanism for ATP production via glycolysis decades
before they understood the mechanism that generates ATP via oxidative phosphorylation. In glycolysis, ATP production is directly linked to the chemical reactions that break glucose down into two molecules of pyruvate. Early studies of oxidative phosphorylation suggested that transport of high-energy electrons down a
cascade of electron acceptors generated the energy for ATP production. By analogy with glycolysis, it was thought that production of ATP would be directly linked
to high-energy chemical intermediates produced during electron transport.
Despite much efort, such compounds were never found. However, a number of
experiments suggested an alternative hypothesis, in which the energy captured
during electron transport was used to pump protons (H+) across the membrane,
generating a gradient of protons that was subsequently used to drive ATP synthesis. he proposed indirect linkage between electron transport and ATP production was known as the chemiosmotic hypothesis for oxidative phosphorylation.
his hypothesis proved to be correct.
Electron transport is carried out by a series of large multiprotein complexes. Early work found that these complexes were in some way associated with the inner mitochondrial membrane, although their function and the nature of their association with the membrane were poorly
understood. Which one of the following observations regarding the
electron-transport complexes would have been most consistent with the
chemiosmotic hypothesis?
A. Eicient electron transport can be detected in puriied preparations of
mitochondrial membranes.
B. Proteins in each electron-transport complex are exposed on both sides of
the inner mitochondrial membrane.
14–85
MCAT STYLE
C. he electron-transport complexes must be embedded in the membrane
to accept electrons.
D. he proteins in each electron-transport complex are exposed only to the
matrix side of the membrane.
In one experiment, mitochondrial membranes were mechanically broken into pieces by subjecting them to high-frequency sound waves.
Which one of the following observations would have been most consistent with the chemiosmotic hypothesis?
A. Adding a source of electrons to the fragmented membranes yielded ATP.
B. Electron-transport complexes were associated with the fragmented
membranes.
C. Fragmented membranes could transport electrons, but could not generate ATP.
D. Fragmented membranes produced protons in response to electron
transport.
14–86
Which of the following experiments would have provided the clearest
proof of the chemiosmotic hypothesis?
A. A decrease in pH inside the mitochondrial matrix could generate ATP in
the complete absence of electron transport.
B. ATP production in intact mitochondria requires both the entire electrontransport chain and the ATP synthase complex.
C. Bacteriorhodopsin, which transports protons across membranes in
response to light, could replace the electron-transport chain.
D. Reconstitution of the electron-transport chain in membranes was suicient to transport protons across the membrane.
14–87
305
Our Colleague, the Late Julian Lewis,
Semaphores the Letter H.
you may like to consider the similarities
and differences between this human
mode of communication and the signaling
networks used by cells.
Chapter 15
307
CHAPTER
Cell Signaling
15
PRINCIPLES OF CELL SIGNALING
IN THIS CHAPTER
TERMS TO LEARN
adaptation (desensitization)
adaptor
contact-dependent signaling
endocrine cell
extracellular signal molecule
GTPase-activating protein (GAP)
GTP-binding protein
guanine nucleotide exchange
factor (GEF)
hormone
interaction domain
ion-channel-coupled receptor
kinase cascade
local mediator
PRINCIPLES OF CELL
SIGNALING
monomeric GTPase
neurotransmitter
paracrine signaling
phosphorylation
primary cilium
protein kinase
protein phosphatase
receptor
scaffold protein
second messenger
serine/threonine kinase
synaptic signaling
tyrosine kinase
DEFINITIONS
Match each deinition below with its term from the list above.
15–1
Protein that binds to a GTP-binding protein and activates it by stimulating release of tightly bound GDP, thereby allowing it to bind GTP.
15–2
General term for a protein that binds a speciic extracellular molecule
(ligand) and initiates a response in the cell.
15–3
Alteration of sensitivity following repeated stimulation, reducing a cell’s
response to that level of stimulus.
15–4
Compact protein module that binds to a particular structural motif in
another protein (or lipid) molecule with which the signaling protein
interacts.
15–5
Short-range cell–cell communication via secreted local mediators that
act on adjacent cells.
15–6
A signal relay chain involving multiple protein kinases, each of which is
activated by phosphorylation and then phosphorylates the next protein
kinase in the sequence.
15–7
Small molecule that is formed in the cytosol, or released into it, in
response to an extracellular signal and that helps to relay the signal to the
interior of the cell.
15–8
Specialized animal cell that secretes a hormone into the blood.
SIGNALING THROUGH
G-PROTEIN-COUPLED
RECEPTORS
SIGNALING THROUGH ENZYMECOUPLED RECEPTORS
ALTERNATIVE SIGNALING
ROUTES IN GENE REGULATION
SIGNALING IN PLANTS
308
Chapter 15: Cell Signaling
15–9
Molecule from outside the cell that communicates the behavior or actions
of other cells in the environment and elicits an appropriate response.
15–10
Enzyme that transfers the terminal phosphate group of ATP to a speciic
amino acid of a target protein.
15–11
Small signal molecule secreted by the presynaptic nerve cell at a chemical synapse to relay the signal to the postsynaptic cell.
15–12
Protein that binds to a GTP-binding protein and inactivates it by stimulating its GTPase activity so that its bound GTP is hydrolyzed to GDP.
15–13
Protein that organizes groups of interacting intracellular signaling proteins into signaling complexes.
15–14
Cell–cell communication in which the signal molecule remains bound to
the signaling cell and only inluences cells that physically touch it.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
15–15
here is no fundamental distinction between signaling molecules that
bind to cell-surface receptors and those that bind to intracellular receptors.
15–16
All second messengers are water-soluble and difuse freely through the
cytosol.
THOUGHT PROBLEMS
15–17
Compare and contrast signaling by neurons to signaling by endocrine
cells. What are the relative advantages of these two mechanisms for cellular communication?
Cells communicate in ways that resemble human communication.
Decide which of the following forms of human communication are analogous to autocrine, paracrine, endocrine, and synaptic signaling by cells.
A. A telephone conversation
B. Talking to people at a cocktail party
C. A radio announcement
D. Talking to yourself
15–18
15–19
How is it that diferent cells can respond in diferent ways to exactly the
same signaling molecule even when they have identical receptors?
15–20
Working out the order in which the individual components in a signaling
pathway act is an essential step in deining the pathway. Imagine that
two protein kinases, PK1 and PK2, act sequentially in a kinase cascade.
When either kinase is completely inactivated, cells do not respond to
the normal extracellular signal. By contrast, cells containing a mutant
form of PK1 that is permanently active respond even in the absence of
an extracellular signal. Doubly mutant cells that contain inactivated PK2
and permanently active PK1 respond in the absence of a signal.
In the normal kinase cascade, does PK1 activate PK2 or does PK2 activate PK1? What outcome would you have predicted for a doubly mutant
cell line with an activating mutation in PK2 and an inactivating mutation
in PK1? Explain your reasoning.
15–21
Why do you suppose that phosphorylation/dephosphorylation, as
opposed to allosteric binding of small molecules, for example, has
evolved to play such a prominent role in switching proteins on and of in
signaling pathways?
PRINCIPLES OF CELL SIGNALING
15–22
15–23
309
he two main classes of molecular switches involve changes in phosphorylation state or changes in guanine nucleotide binding. Comment on
the following statement. “In the regulation of molecular switches, protein kinases and guanine nucleotide exchange factors (GEFs) always turn
proteins on, whereas protein phosphatases and GTPase-activating proteins (GAPs) always turn proteins of.”
Consider a signaling pathway that proceeds through three protein
kinases that are sequentially activated by phosphorylation. In one case,
the kinases are held in a signaling complex by a scafold protein; in the
other, the kinases are freely difusing (Figure 15–1). Discuss the properties of these two types of organization in terms of signal ampliication,
speed, and potential for cross-talk between signaling pathways.
Proteins in signaling pathways use a variety of binding domains to
assemble into signaling complexes. Match the following domains with
their binding targets. (A binding target can be used more than once.)
A. PH domain
1. phosphorylated tyrosines
B. PTB domain
2. proline-rich sequences
C. SH2 domain
3. phosphorylated inositol phospholipids
D. SH3 domain
15–24
15–25
Describe three ways in which a gradual increase in an extracellular signal
can be sharpened by the target cell to produce an abrupt or nearly all-ornone response.
CALCULATIONS
15–26
Suppose that the circulating concentration of hormone is 10–10 M and
the Kd for binding to its receptor is 10–8 M. What fraction of the receptors will have hormone bound? If a meaningful physiological response
occurs when 50% of the receptors have bound a hormone molecule, how
much will the concentration of hormone have to rise to elicit a response?
Recall that the fraction of receptors (R) bound to hormone (H) to form a
receptor–hormone complex (R–H) is [R–H]/([R] + [R–H]) = [R–H]/[R]TOT
= [H]/([H] + Kd).
15–27
Radioimmunoassay (RIA) is a powerful tool for quantifying virtually
any substance of biological interest because it is sensitive, accurate, and
fast. RIA technology arose from studies on adult-onset diabetes. Some
patients had antibodies with high ainity for insulin, and RIA was developed as a method to distinguish free insulin from antibody-bound insulin.
How can high-ainity antibodies be exploited to measure low concentrations of insulin? When a small amount of insulin-speciic antiserum is mixed with an equally small amount of very highly radioactive
insulin, some binds and some remains free according to the equilibrium.
Insulin (I) + Antibody (A)
Insulin–Antibody (I–A) Complex
When increasing amounts of unlabeled insulin are added to a ixed
amount of labeled insulin and anti-insulin antibody, the ratio of bound
to free radioactive insulin decreases as expected from the equilibrium
expression. If the concentration of the unlabeled insulin is known, then
the resulting curve serves as a calibration against which other unknown
samples can be compared (Figure 15–2).
You have three samples of insulin whose concentrations are unknown.
When mixed with the same amount of radioactive insulin and anti-insulin antibody used in Figure 15–2, the three samples gave the following
ratios of bound to free insulin:
CYTOSOL
1
1
2
3
2
3
Figure 15–1 A protein kinase cascade
organized by a scaffolding protein or
composed of freely diffusing components
(Problem 15–23).
Chapter 15: Cell Signaling
310
Figure 15–2 Calibration curve
for radioimmunoassay of insulin
(Problem 15–27).
1.4
bound/free (ratio)
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0
5
10
unlabeled insulin (pg/mL)
15
Sample 1
0.67
Sample 2
0.31
Sample 3
0.46
A. What is the concentration of insulin in each of these unknown samples?
B. What portion of the standard curve is the most accurate, and why?
C. If the antibodies were raised against pig insulin, which is similar but not
identical to human insulin, would the assay still be valid for measuring
human insulin concentrations?
Two intracellular molecules, A and B, are normally synthesized at a constant rate of 1000 molecules per second per cell. Each molecule of A survives an average of 100 seconds, while each molecule of B survives an
average of 10 seconds.
A. How many molecules of A and B will a cell contain?
B. If the rates of synthesis of both A and B were suddenly increased 10-fold
to 10,000 molecules per second—without any change in their average
lifetimes—how many molecules of A and B would be present after 1 second?
C. Which molecule would be preferred for rapid signaling? Explain your
answer.
15–28
DATA HANDLING
15–29
he cellular slime mold Dictyostelium discoideum is a eukaryote that
lives on the forest loor as independent motile cells called amoebae,
which feed on bacteria and yeast. When their food supply is exhausted,
the amoebae stop dividing and gather together to form tiny, multicellular, wormlike structures, which crawl about as slugs. How do individual
amoebae know when to stop dividing and how to ind their way into a
common aggregate? A set of classic experiments investigated this phenomenon more than half a century ago.
Amoebae aggregate when placed on a glass coverslip under water,
provided that simple salts are present. he center of the aggregation pattern can be removed with a pipette and placed in a ield of fresh amoebae,
which immediately start streaming toward it. hus, the center is emitting
some sort of attractive signal. Four experiments were designed to determine the nature of the signal. In each, an existing center of aggregation
was used as the source of the signal and previously unexposed amoebae
served as the target cells. he arrangements of aggregation centers and
test amoebae at the beginning and end of the experiments are shown in
Figure 15–3.
Do these results show that Dictyostelium discoideum aggregates
through the action of a secreted chemical signal? Explain your reasoning.
15–30
he nicotinic acetylcholine receptor is a neurotransmitter-dependent ion
channel, which is composed of four types of subunit. Phosphorylation of
PRINCIPLES OF CELL SIGNALING
311
FINAL ARRANGEMENT
INITIAL ARRANGEMENT
actively signaling
aggregation center
(A)
lower center forms
at random location
glass
coverslip
layer of amoebae
glass
coverslip
semipermeable
membrane
aggregation center
placed at edge
(B)
lower amoebae stream
around the edge
lower center forms
exactly below upper
center
(C)
(D) (top view)
gentle
stream of
medium
across
coverslip
the receptor by protein kinase A attaches one phosphate to the γ subunit and one phosphate to the δ subunit. Fully phosphorylated receptors
desensitize much more rapidly than unmodiied receptors. To study this
process in detail, you phosphorylate two preparations of receptor to different extents (0.8 mole phosphate/mole receptor and 1.2 mole phosphate/mole receptor) and measure desensitization over several seconds
(Figure 15–4). Both preparations behave as if they contain a mixture of
receptors; one form that is rapidly desensitized (the initial steep portion
of the curves) and another form that is desensitized at the same rate as
the untreated receptor.
A. Assuming that the γ and δ subunits are independently phosphorylated
at equal rates, calculate the percentage of receptors that carry zero, one,
and two phosphates per receptor at the two extents of phosphorylation.
B. Do these data suggest that desensitization requires one phosphate or two
phosphates per receptor? If you decide that desensitization requires only
one phosphate, indicate whether the phosphate has to be on one speciic
subunit or can be on either of the subunits.
amoebae downstream
of center stream to
join it; upstream
amoebae ignore the
center
Figure 15–3 Four experiments to study
the nature of the attractive signal
generated by aggregation centers
(Problem 15–29).
MEDICAL LINKS
Surgeons use succinylcholine, which is an acetylcholine analog, as a
muscle relaxant. Care must be taken because some individuals recover
abnormally slowly from this paralysis, with life-threatening consequences. Such individuals are deicient in an enzyme called pseudocholinesterase, which is normally present in the blood, where it slowly
inactivates succinylcholine by hydrolysis to succinate and choline.
If succinylcholine is an analog of acetylcholine, why do you think it
causes muscles to relax and not contract as they do in the presence of
acetylcholine?
Figure 15–4 Desensitization rates for untreated acetylcholine receptor
and two preparations of phosphorylated receptor (Problem 15–30). Red
squares represent untreated receptors; blue squares represent receptors
with 0.8 mole phosphate/mole receptor; and brown triangles represent
receptors with 1.2 mole phosphate/mole receptor. Arrows indicate
the fractions of the phosphorylated preparations that behaved like the
untreated receptor.
100
activity remaining (%)
15–31
36
18
10
1
0
2
4
preincubation time
(seconds)
6
312
Chapter 15: Cell Signaling
SIGNALING THROUGH G-PROTEIN-COUPLED
RECEPTORS
TERMS TO LEARN
adenylyl cyclase
GPCR kinase (GRK)
arrestin
inhibitory G protein (Gi)
Ca2+/calmodulin-dependent
inositol phospholipid signaling pathway
kinase (CaM-kinase)
inositol 1,4,5-trisphosphate (IP3)
calmodulin
IP3 receptor
CaM-kinase II
nitric oxide (NO)
cone photoreceptor
NO synthase
CRE-binding (CREB) protein
olfactory receptor
cyclic AMP (cAMP)
phospholipase C-β (PLCβ)
cyclic-AMP-dependent protein phosphatidylinositol 4,5-bisphosphate
kinase (PKA)
(PI(4,5)P2)
cyclic AMP phosphodiesterase protein kinase C (PKC)
cyclic GMP
regulator of G protein signaling (RGS)
cyclic GMP phosphodiesterase rhodopsin
diacylglycerol
rod photoreceptor (rod)
Gq
ryanodine receptor
G-protein-coupled receptor
stimulatory G protein (Gs)
(GPCR)
trimeric GTP-binding protein (G protein)
DEFINITIONS
Match each deinition below with its term from the list above.
15–32
G protein that activates adenylyl cyclase and thereby increases cyclic
AMP concentration.
15–33
Protein composed of three subunits, one of which is activated by the
binding of GTP.
15–34
Ubiquitous calcium-binding protein whose interactions with other proteins are governed by changes in intracellular Ca2+ concentration.
15–35
Enzyme that hydrolyzes cyclic AMP to adenosine 5ʹ-monophosphate (5ʹAMP).
15–36
Cell-surface receptor that associates with an intracellular G protein upon
activation by an extracellular ligand.
15–37
Enzyme that participates in desensitization of GPCRs by phosphorylating them after they have been activated by ligand binding.
15–38
Ca2+-release channel in the ER membrane that is activated by Ca2+ binding in the absence of IP3.
15–39
Enzyme bound to the cytoplasmic surface of the plasma membrane that
converts membrane PI(4,5)P2 to diacylglycerol and IP3.
15–40
Protein that is an α-subunit-speciic GTPase-activating protein (GAP).
15–41
Second messenger that is released from a phospholipid in the plasma
membrane and difuses to the ER, where it opens Ca2+-release channels.
15–42
Enzyme that phosphorylates target proteins in response to a rise in intracellular cyclic AMP.
15–43
A Ca2+-dependent protein kinase that is activated by diacylglycerol.
15–44
Light-sensitive GPCR in rod photoreceptor cells of the retina.
15–45
Protein kinase whose activity is regulated by the binding of Ca2+-activated
SIGNALING THROUGH G-PROTEIN-COUPLED RECEPTORS
calmodulin, and which indirectly mediates the efects of Ca2+ by phosphorylation of other proteins.
15–46
Protein that binds to the cyclic AMP response elements found in the regulatory region of many genes activated by cyclic AMP.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
15–47
Diferent isoforms of protein kinase A in diferent cell types explain why
the efects of cyclic AMP vary depending on the target cell.
15–48
he activity of any protein regulated by phosphorylation depends on the
balance at any instant between the activities of the kinases that phosphorylate it and the phosphatases that dephosphorylate it.
15–49
Most intracellular signaling pathways provide multiple opportunities for
amplifying a response to an extracellular signal.
THOUGHT PROBLEMS
15–50
GPCRs activate G proteins by reducing the strength of GDP binding,
allowing GDP to dissociate and GTP, which is present at much higher
concentrations, to bind. How do you suppose the activity of a G protein
would be afected by a mutation that caused its ainity for GDP to be
reduced without signiicantly changing its ainity for GTP?
15–51
When adrenaline (epinephrine) binds to adrenergic receptors on the
surface of a muscle cell, it activates a G protein, initiating a signaling
pathway that results in breakdown of muscle glycogen. How would you
expect glycogen breakdown to be afected if muscle cells were injected
with a nonhydrolyzable analog of GTP, which can’t be converted to GDP?
Consider what would happen in the absence of adrenaline and after a
brief exposure to it.
15–52
Should RGS (regulator of G protein signaling) proteins be classiied as
GEFs (guanine nucleotide exchange factors) or GAPs (GTPase-activating
proteins)? Explain what role this activity plays in modulating G-proteinmediated responses in animals and yeasts.
15–53
What is “cyclic” about cyclic AMP?
15–54
Explain why cyclic AMP must be broken down rapidly in a cell to allow
rapid signaling.
15–55
You are trying to purify adenylyl cyclase from brain. he assay is based
on the conversion of α-32P-ATP to cAMP. You can easily detect activity
in crude brain homogenates stimulated by isoproterenol, which binds to
β-adrenergic receptors, but the enzyme loses activity when low-molecular-weight cofactors are removed by dialysis. What single molecule do
you think you could add back to the dialyzed homogenate to restore
activity?
15–56
Propose speciic types of mutations in the gene for the regulatory subunit
of cyclic-AMP-dependent protein kinase (PKA) that could lead to either a
permanently active PKA, or to a permanently inactive PKA.
15–57
Why do you suppose cells use Ca2+ (intracellular concentration 10–7 M)
for signaling rather than the more abundant Na+ (intracellular concentration 10–3 M)?
15–58
EGTA chelates Ca2+ with high ainity and speciicity. How would microinjection of EGTA afect glucagon-triggered breakdown of glycogen in
313
Chapter 15: Cell Signaling
314
cAMP
Ca2+
Figure 15–5 Integration of cyclic-AMPdependent and Ca2+-dependent signaling
pathways by phosphorylase kinase in
liver and muscle cells (Problem 15–59).
G1P is glucose 1-phosphate, the product
of phosphorylase kinase, which uses
phosphate to cleave glucose units from
glycogen.
P
P
Ca2+
Ca2+
phosphorylase
kinase
P
glycogen
phosphorylase
GLYCOGEN
G1P
liver? How would it afect vasopressin-triggered breakdown of glycogen
in liver?
15–59
Phosphorylase kinase integrates signals from the cyclic-AMP-dependent
and Ca2+-dependent signaling pathways that control glycogen breakdown in liver and muscle cells (Figure 15–5). Phosphorylase kinase is
composed of four subunits. One is the protein kinase that catalyzes the
addition of phosphate to glycogen phosphorylase to activate it for glycogen breakdown. he other three subunits are regulatory proteins that
control the activity of the catalytic subunit. Two contain sites for phosphorylation by PKA, which is activated by cyclic AMP. he remaining
subunit is calmodulin, which binds Ca2+ when the cytosolic Ca2+ concentration rises. he regulatory subunits control the equilibrium between
the active and inactive conformations of the catalytic subunit, with each
phosphate and Ca2+ nudging the equilibrium toward the active formation. How does this arrangement allow phosphorylase kinase to serve its
role as an integrator protein for the multiple pathways that stimulate glycogen breakdown?
15–60
CaM-kinase II is a remarkable molecular memory device. How does
CaM-kinase II “remember” its exposure to Ca2+/calmodulin and why
does it eventually “forget”?
he outer segments of rod photoreceptor cells can be broken of, isolated,
and used to study the efects of small molecules on visual transduction
because the broken end of each segment remains unsealed. How would
you expect the visual response to be afected by the following additions?
A. An inhibitor of cyclic GMP phosphodiesterase.
B. A nonhydrolyzable analog of GTP.
C. An inhibitor of rhodopsin-speciic kinase.
15–61
15–62
A rise in the cyclic GMP levels in smooth muscle cells causes relaxation
of the blood vessels in the penis, resulting in an erection. Explain how the
natural signal molecule NO and the drug Viagra® produce an increase in
cyclic GMP.
15–63
In muscle cells, adrenaline binds to the β-adrenergic receptor to initiate a signaling cascade that leads to the breakdown of glycogen (Figure
15–6). At what points in this pathway is the signal ampliied?
15–64
A critical feature of all signaling cascades is that they must be turned of
rapidly when the extracellular signal is removed. Examine the signaling
cascade in Figure 15–6. Describe how each component of this signaling
pathway is returned to its inactive state when adrenaline is removed.
SIGNALING THROUGH G-PROTEIN-COUPLED RECEPTORS
adrenaline
α
Figure 15–6 Signaling cascade for
activation of glycogen breakdown by
adrenaline in muscle cells (Problem 15–63).
G1P is glucose 1-phosphate; cAMP bound
to the regulatory subunits of PkA is shown
as red balls.
adenylyl
cyclase
CYTOSOL
α
β
GDP
315
γ
GTP
ATP
dissociated
R-subunit
of PKA
cAMP
inactive PKA
active
PKA
P
P
phosphorylase
kinase
P
glycogen
phosphorylase
GLYCOGEN
G1P
CALCULATIONS
In a classic paper, the number of β-adrenergic receptors on the membranes of frog erythrocytes was determined by using a competitive inhibitor of adrenaline, alprenolol, which binds to the receptors 500 times
more tightly than adrenaline. hese receptors normally bind adrenaline
and stimulate adenylyl cyclase activity. Labeled alprenolol was mixed
with erythrocyte membranes, left for 10 minutes at 37°C, and then the
membranes were pelleted by centrifugation and the radioactivity in the
pellet was measured. he experiment was done in two ways. he binding
of increasing amounts of 3H-alprenolol to a ixed amount of erythrocyte
membranes was measured to determine total binding. he experiment
was repeated in the presence of a vast excess of unlabeled alprenolol to
measure nonspeciic binding. he results are shown in Figure 15–7.
A. On Figure 15–7, sketch the curve for speciic binding of alprenolol to
β-adrenergic receptors. Has alprenolol binding to the receptors reached
saturation?
B. Assuming that one molecule of alprenolol binds per receptor, calculate
the number of β-adrenergic receptors on the membrane of a frog erythrocyte. he speciic activity of the labeled alprenolol is 1 ×1013 cpm/
mmol, and there are 8 ×108 frog erythrocytes per milligram of membrane
protein.
15–65
bound
(cpm/mg x 103)
In visual transduction, one activated rhodopsin molecule leads to the
hydrolysis of 5 ×105 cyclic GMP molecules per second. One stage in this
enormous signal ampliication is achieved by cyclic GMP phosphodiesterase, which hydrolyzes 1000 molecules of cyclic GMP per second. he
additional factor of 500 could arise because one activated rhodopsin activates 500 transducin (Gt) molecules, or because one activated transducin
activates 500 cyclic GMP phosphodiesterases, or through a combination
of both efects. One experiment to address this question measured the
3H-alprenolol
15–66
40
total
30
20
nonspecific
10
0
0
2
4
6
3H-alprenolol
8
10
(nM)
Figure 15–7 Binding of 3H-alprenolol to frog
erythrocyte membranes (Problem 15–65).
Chapter 15: Cell Signaling
316
Figure 15–8 Binding of GppNp to rod-cell
membranes as a function of the fraction
of activated rhodopsin (Problem 15–66).
Background binding of GppNp to rodcell membranes in the dark has been
subtracted from the values shown.
GppNp bound
(mmol/mole of rhodopsin)
12
10
8
6
5.5 mmol/mole
4
1.1 × 10–3 %
activated
2
0
10-5
10-4
10-3
10-2
10-1
activated rhodopsin (%)
amount of GppNp (a nonhydrolyzable analog of GTP) that is bound by
transducin in the presence of diferent amounts of activated rhodopsin.
As indicated in Figure 15–8, 5.5 mmol of GppNp were bound per mole of
total rhodopsin when 0.0011% of the rhodopsin was activated.
A. Assuming that each transducin molecule binds one molecule of GppNp,
calculate the number of transducin molecules that are activated by each
activated rhodopsin molecule. Which mechanism of ampliication does
this measurement support?
B. Binding studies have shown that transducin-GDP has a high ainity for
activated rhodopsin and that transducin-GTP has a low ainity; conversely, transducin-GTP has a high ainity and transducin-GDP has a
low ainity for cyclic GMP phosphodiesterase. Are these ainities consistent with the mechanism of ampliication you deduced from the above
experiment? Explain your reasoning.
DATA HANDLING
he mating behavior of yeast depends on signaling peptides termed
pheromones that bind to pheromone GPCRs (Figure 15–9). When the
α-factor pheromone binds to a wild-type yeast cell, it blocks cell-cycle
progression, arresting proliferation until a mating partner is found. Yeast
mutants with defects in one or more of the components of the G protein
have characteristic phenotypes in the absence and in the presence of the
α-factor pheromone (Table 15–1). Strains with defects in any of these
genes cannot undergo the mating response and are therefore termed
sterile.
A. Based on genetic analysis of the yeast mutants, decide which component
of the G protein normally transmits the mating signal to the downstream
efector molecules.
B. Predict the proliferation and mating phenotypes in the absence and
presence of the α-factor pheromone of strains with the following mutant
G protein α subunits:
15–67
α-factor
EXTRACELLULAR
GTP
α-factor
GDP
α
β
receptor
γ
GDP
Figure 15–9 α-Factor pheromone signaling via α-factor GPCR and G protein (Problem 15–67).
α
β
GTP
γ
SIGNALING THROUGH G-PROTEIN-COUPLED RECEPTORS
317
TABLE 15–1 Mating phenotypes of various mutant and nonmutant strains of
yeast (Problem 15–67).
Mutation
Phenotype
Minus α factor
Plus α factor
None (wild type)
Normal proliferation
Arrested proliferation, mating
response
α subunit deleted
Arrested proliferation
Arrested proliferation, sterile
β subunit deleted
Normal proliferation
Normal proliferation, sterile
γ subunit deleted
Normal proliferation
Normal proliferation, sterile
α and β deleted
Normal proliferation
Normal proliferation, sterile
α and γ deleted
Normal proliferation
Normal proliferation, sterile
β and γ deleted
Normal proliferation
Normal proliferation, sterile
1. An α subunit that can bind GTP but cannot hydrolyze it.
2. An α subunit with an altered N-terminus to which the fatty acid myristoylate cannot be added, thereby preventing its localization to the plasma
membrane.
3. An α subunit that cannot bind to the activated pheromone receptor.
A particularly graphic illustration of the subtle, yet important, role of
cyclic AMP in the whole organism comes from studies of the fruit ly Drosophila melanogaster. In search of the gene for cyclic AMP phosphodiesterase, one laboratory measured enzyme levels in lies with chromosomal duplications or deletions and found consistent alterations in lies
with mutations involving bands 3D3 and 3D4 on the X chromosome.
Duplications in this region have about 1.5 times the normal activity of
the enzyme; deletions have about half the normal activity.
An independent laboratory at the same institution was led to the
same chromosomal region through work on behavioral mutants of fruit
lies. he researchers had developed a learning test in which lies were
presented with two metallic grids, one of which was electriied. If the
electriied grid was painted with a strong-smelling chemical, normal lies
quickly learned to avoid it, even when it was no longer electriied. he
mutant lies, on the other hand, never learned to avoid the smelly grid;
they were aptly called Dunce mutants. he Dunce mutation was mapped
genetically to bands 3D3 and 3D4.
Is the learning defect really due to lack of cyclic AMP phosphodiesterase or are the responsible genes simply closely linked? Further experiments showed that the level of cyclic AMP in Dunce lies was 1.6 times
higher than in normal lies. Furthermore, sucrose-gradient analysis of
homogenates of Dunce and normal lies revealed two cyclic AMP phosphodiesterase activities, one of which was missing in Dunce lies (Figure
15–10).
A. Why do Dunce lies have higher levels of cyclic AMP than normal lies?
B. Explain why homozygous (both chromosomes afected) duplications of
the nonmutant Dunce gene cause cyclic AMP phosphodiesterase levels
to be elevated 1.5-fold and why homozygous deletions of the gene reduce
enzyme activity to half the normal value.
C. What would you predict would be the efect of cafeine, a phosphodiesterase inhibitor, on the learning performance of normal lies?
15–68
phosphodiesterase
activity
normal
Dunce
top
bottom
Figure 15–10 Sucrose-gradient analysis
of cyclic AMP phosphodiesterase activity
in homogenates of normal and Dunce
lies (Problem 15–68).
Chapter 15: Cell Signaling
318
(A)
(B)
(C)
(D)
+
+
+
+
intact
cell
buffer
buffer
+ GTP
buffer
Figure 15–11 Experimental set-up and
typical results of patch-clamp analysis of
k+-channel activation by acetylcholine
(Problem 15–69). The buffer is a salts
solution that does not contain nucleotides
or Ca2+. In all these experiments,
acetylcholine is present inside the pipet,
as indicated by the plus sign. The current
through the membrane is measured in
picoamps (pA). In (C), the GTP is added
to the buffer.
1 pA
200 seconds
Acetylcholine acts on muscarinic GPCRs in the heart to open K+ channels, thereby slowing the heart rate. his process can be directly studied
using the inside-out membrane patch-clamp technique. he external
surface of the membrane is in contact with the solution in the bore of
the pipet, and the cytoplasmic surface faces outward and can be exposed
readily to a variety of solutions (Figure 15–11). Receptors, G proteins,
and K+ channels remain associated with the membrane patch.
When acetylcholine is added to a pipet with a whole cell attached, K+
channels open as indicated by the low of current (Figure 15–11A). Under
similar circumstances with a patch of membrane inserted into a bufered
salts solution, no current lows (Figure 15–11B). When GTP is added to
the bufer, current resumes (Figure 15–11C). Subsequent removal of GTP
stops the current (Figure 15–11D). he results of several similar experiments to test the efects of diferent combinations of components are
summarized in Table 15–2.
A. Why do you think it is that Gβγ activated the channel when the complete
G protein did not? Is the active component of the G protein in this system
the same as the one that activates adenylyl cyclase in other cells?
B. Addition of GppNp (a nonhydrolyzable analog of GTP) causes the K+
channel to open in the absence of acetylcholine (Table 15–2, line 4). he
low of current, however, rose very slowly and reached its maximum only
after a minute (compare with the immediate rise in Figure 15–11A and
C). How do you suppose GppNp causes the channels to open slowly in
the absence of acetylcholine?
C. To the extent that these experiments allow, draw a scheme for the activation of K+ channels in heart cells in response to acetylcholine.
15–69
TABLE 15–2 Responses of K+ channel to various experimental manipulations (Problem 15–69).
Additions
Acetylcholine
Small molecules
G-protein components
k+ channel
1
+
none
none
closed
2
+
GTP
none
open
3
–
GTP
none
closed
4
–
GppNp
none
open
5
–
none
G protein
closed
6
–
none
Gα
closed
7
–
none
Gβγ
open
8
–
none
boiled G protein
closed
SIGNALING THROUGH G-PROTEIN-COUPLED RECEPTORS
319
MEDICAL LINKS
15–70
During a marathon, runners draw heavily on their internal reserves of
glycogen (carbohydrate) and triglycerides (fat) to fuel muscle contraction. Initially, energy is derived mostly from carbohydrates, with increasing amounts of fat being used as the race progresses. If runners use up
their muscle glycogen reserves before they inish the race, they hit what
is known as “the wall,” a point of diminished performance that arises
because fatty acids from triglyceride breakdown cannot be delivered to
the muscles quickly enough to sustain maximum efort. One trick that
marathon runners use to avoid the wall is to drink a cup of strong black
cofee an hour or so before the race begins. Cofee contains cafeine,
which is an inhibitor of cyclic AMP phosphodiesterase. How do you suppose inhibition of this enzyme helps them avoid the wall?
15–71
Patients with Oguchi’s disease have an inherited form of night blindness.
After a lash of bright light, these individuals recover their night vision
(become dark adapted) very slowly. Night vision depends almost entirely
on the visual responses of rod photoreceptor cells. What aspect of the
visual response in these patients’ rod cells do you suppose is defective?
What genes, when defective, might give rise to Oguchi’s disease?
he primary role of platelets is to control blood clotting. When they
encounter the exposed basement membrane (collagen ibers) of a damaged blood vessel or a newly forming ibrin clot, they change their shape
from round to spiky and stick to the damaged area. At the same time, they
begin to secrete serotonin and ATP, which accelerate similar changes
in newly arriving platelets, leading to the rapid formation of a clot. he
platelet response is regulated by protein phosphorylation. Signiicantly,
platelets contain high levels of two protein kinases: PKC, which initiates
serotonin release, and myosin light-chain kinase, which mediates the
change in shape.
When platelets are stimulated with thrombin, the light chain of
myosin and an unknown protein of 40,000 daltons are phosphorylated.
When platelets are treated with a calcium ionophore, which increases
membrane permeability to Ca2+, only the myosin light chain is phosphorylated; when they are treated with diacylglycerol, only the 40-kD
protein is phosphorylated. Experiments using a range of concentrations of diacylglycerol in the presence or absence of calcium ionophore
show that the extent of phosphorylation of the 40-kD protein depends
only on the concentration of diacylglycerol (Figure 15–12A). Serotonin
release, however, depends on diacylglycerol and the calcium ionophore
(Figure 15–12B).
A. Based on these experimental observations, describe the normal sequence
of molecular events that leads to phosphorylation of the myosin light
15–72
(B) SEROTONIN RELEASE
100
serotonin release
(% maximum)
40-kD protein phosphorylation
(% maximum)
(A) PHOSPHORYLATION
100
50
0
0
10
20
diacylglycerol (µg/mL)
+ ionophore
50
– ionophore
0
0
10
20
diacylglycerol (µg/mL)
Figure 15–12 Treatment of platelets with
calcium ionophore and diacylglycerol
(Problem 15–72). (A) Effects on
phosphorylation of the 40-kD protein.
(B) Effects on serotonin release. Red
circles indicate the presence of calcium
ionophore and blue circles indicate its
absence.
Chapter 15: Cell Signaling
320
chain and the 40-kD protein. Indicate how the calcium ionophore and
diacylglycerol treatments interact with the normal sequence of events.
B. Why do you think serotonin release requires both calcium ionophore
and diacylglycerol?
SIGNALING THROUGH ENZYME-COUPLED
RECEPTORS
TERMS TO LEARN
Akt
Cdc42
cytokine receptor
cytoplasmic tyrosine kinase
enzyme-coupled receptor
ephrins
focal adhesion kinase (FAK)
JAK–STAT signaling pathway
Janus kinase (JAK)
MAP kinase module
mTOR
phosphoinositide
phosphoinositide 3-kinase
(PI 3-kinase)
phospholipase C-γ (PLCγ)
PI-3-kinase–Akt pathway
pleckstrin homology (PH) domain
protein tyrosine phosphatase
Rac
Ras
Ras-GAP
Ras-GEF
Ras–MAP-kinase signaling pathway
Ras superfamily
receptor serine/threonine kinase
receptor tyrosine kinase (RTK)
Rheb
Rho
Rho family
SH2 domain
Smad family
Src family
STATs
TOR
transforming growth factor-β (TGFβ)
superfamily
tyrosine-kinase-associated receptor
DEFINITIONS
Match each deinition below with its term from the list above.
15–73
he largest class of cell-surface-bound extracellular signal proteins.
15–74
Large family of structurally related, secreted, dimeric proteins that act as
hormones and local mediators to control a wide range of biological functions in all animals.
15–75
Cell-surface receptor that when activated by ligand binding adds phosphates from ATP to tyrosine side chains in its own cytoplasmic domain.
15–76
he founding member of a superfamily of monomeric GTPases that help
to relay signals from cell-surface receptors to the nucleus.
15–77
A group of monomeric GTPases that regulate both the actin and microtubule cytoskeletons.
15–78
Cytoplasmic tyrosine kinase present at cell–matrix junctions in association with the cytoplasmic tails of integrins.
15–79
A kinase that is involved in intracellular signaling pathways activated by
cell-surface receptors and that phosphorylates inositol phospholipids at
the 3 position of the inositol ring.
15–80
Cell-surface receptor in which the cytoplasmic domain either has enzymatic activity itself or is associated with an intracellular enzyme.
15–81
Cell-surface receptor that activates a tyrosine kinase that is noncovalently bound to the receptor.
15–82
A three-component signaling module used in various signaling pathways
in eukaryotic cells.
SIGNALING THROUGH ENZYME-COUPLED RECEPTORS
15–83
One of several intracellular signaling pathways that leads from cell-surface receptors to the nucleus, it is distinguished by providing one of the
more direct routes.
15–84
Protein domain found in intracellular signaling proteins by which they
bind to inositol phospholipids phosphorylated by PI 3-kinase.
15–85
A protein domain that is homologous to a region in Src, is present in
many proteins, and binds to a short amino acid sequence containing a
phosphotyrosine.
15–86
A crucial signaling protein in the PI-3-kinase–Akt signaling pathway, so
named because it is the target of rapamycin.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
15–87
Binding of extracellular ligands to receptor tyrosine kinases (RTKs) activates the intracellular catalytic domain by propagating a conformational
change across the lipid bilayer through a single transmembrane α helix.
15–88
PI 3-kinase phosphorylates the inositol head groups of phospholipids at
the 3 position of the ring so that they can be cleaved by phospholipase C
to produce IP3.
15–89
Protein tyrosine phosphatases display exquisite speciicity for their substrates, unlike most serine/threonine protein phosphatases, which have
rather broad speciicity.
THOUGHT PROBLEMS
15–90
Antibodies are Y-shaped molecules that carry two identical binding sites.
Imagine that you have obtained an antibody that is speciic for the extracellular domain of a receptor tyrosine kinase. If cells were exposed to the
antibody, would you expect the receptor tyrosine kinase to be activated,
inactivated, or unafected? Explain your reasoning.
Genes encoding mutant forms of a receptor tyrosine kinase can be introduced into cells that express the normal receptor from their own genes.
If the mutant genes are expressed at considerably higher levels than the
normal genes, what will be the consequences for receptor-mediated signaling of introducing genes for the following mutant receptors?
A. A mutant receptor tyrosine kinase that lacks its extracellular domain.
B. A mutant receptor tyrosine kinase that lacks its intracellular domain.
15–91
he SH3 domain, which comprises about 60 amino acids, recognizes
and binds to structural motifs in other proteins. he motif recognized
by SH3 domains was found by constructing a fusion protein between an
SH3 domain and glutathione-S-transferase (GST). GST fusions allow for
easy puriication using a glutathione ainity column, which binds GST
speciically. After tagging the puriied GST–SH3 protein with biotin to
make it easy to detect, it was used to screen ilters containing E. coli colonies expressing a cDNA library. Two diferent clones were identiied that
bound to the SH3 domain: in both cases, binding was shown to occur at
short proline-rich sequences.
A. Could you use biotin-tagged GST–SH2 proteins in the same way to ind
cDNAs for proteins that bind to SH2 domains? Why or why not?
B. Many proteins bind to short strings of amino acids in other proteins. How
do you think these kinds of interactions difer from the kinds of interactions found between the protein subunits of multisubunit enzymes?
15–92
321
322
Chapter 15: Cell Signaling
15–93
he Ras protein functions as a molecular switch that is turned on by a
guanine nucleotide exchange factor (GEF) that causes it to bind GTP. A
GTPase-activating protein (GAP) turns the switch of by inducing Ras to
hydrolyze its bound GTP to GDP much more rapidly than in the absence
of the GAP. hus, Ras works like a light switch that one person turns
on and another turns of. In a cell line that lacks the Ras-speciic GAP,
what abnormalities in Ras activity, if any, would you expect to ind in the
absence of extracellular signals, and in their presence?
15–94
What are the similarities and diferences between the reactions that lead
to the activation of G proteins and those that lead to the activation of Ras?
15–95
In principle, the activated, GTP-bound form of Ras could be increased
by activating a guanine nucleotide exchange factor (GEF) or by inactivating a GTPase-activating protein (GAP). Why do you suppose that Rasmediated signaling pathways always increase Ras-GTP by activating a
GEF rather than inactivating a GAP?
15–96
A single amino acid change in Ras eliminates its ability to hydrolyze GTP,
even in the presence of a GTPase-activating protein (GAP). Roughly
30% of human cancers have this change in Ras. You have just identiied
a small molecule that prevents the dimerization of a receptor tyrosine
kinase that signals via Ras. Would you expect this molecule to be efective in the treatment of cancers that express this common, mutant form
of Ras? Why or why not?
(A) RECEPTORS
1
What does autophosphorylation mean? When a receptor tyrosine kinase
binds its ligand and forms a dimer, do the individual receptor molecules
phosphorylate themselves or does one receptor cross-phosphorylate the
other, and vice versa? To investigate this question, you’ve constructed
genes for three forms of a receptor tyrosine kinase: the normal form with
an active kinase domain and three sites of phosphorylation; a large form
that carries an inactivating point mutation in the kinase domain but
retains the three phosphorylation sites; and a short version that has an
active kinase domain but is lacking the sites of phosphorylation (Figure
15–13A). You express the genes singly and in combination in a cell line
that lacks this receptor tyrosine kinase, and then break open the cells and
add the ligand for the receptor in the presence of radioactive ATP. You
immunoprecipitate the receptors and analyze them for expression levels
by staining for protein (Figure 15–13B) and for phosphorylation by autoradiography (Figure 15–13C).
A. What results would you expect on the autoradiograph if individual receptors only phosphorylated themselves?
B. What would you expect if receptors cross-phosphorylated each other?
C. Which model for autophosphorylation do your data support?
15–97
15–98
When activated, the platelet-derived growth factor (PDGF) receptor
phosphorylates itself on multiple tyrosines. hese phosphorylated tyrosines serve as assembly sites for several SH2-domain-containing proteins
that include phospholipase C-γ (PLCγ), a Ras-speciic GTPase-activating protein (GAP), a subunit of phosphoinositide 3-kinase (PI3K), and
a phosphotyrosine phosphatase (PTP) (Figure 15–14). PDGF binding
stimulates several changes in the target cell, one of which is an increase
in DNA synthesis, as measured by incorporation of radioactive thymidine or bromodeoxyuridine into DNA.
To determine which of the bound proteins is responsible for activation of DNA synthesis, you construct several mutant genes for the
PDGF receptor that retain individual or combinations of tyrosine
3
–
+
+
–
CYTOSOL
Y
Y
Y
Y
Y
Y
kinase
P sites
+
+
(B) PROTEIN GEL
1
2
3
1+2 1+3 2+3
2
1
3
(C) RADIOACTIVITY
1
DATA HANDLING
2
2
3
1+2 1+3 2+3
2
1
3
Figure 15–13 Analysis of
autophosphorylation (Problem 15–97).
(A) Normal and mutant receptor tyrosine
kinases. P sites refers to the sites
of phosphorylation. (B) Expression
of receptor tyrosine kinases.
(C) Phosphorylation of receptor
tyrosine kinases.
SIGNALING THROUGH ENZYME-COUPLED RECEPTORS
323
Figure 15–14 The signaling complex
assembled on the PDGF receptor
(Problem 15–98). Numbers refer to the
positions of the phosphorylated amino
acids in the sequence of the PDGF
receptor.
PDGF
CYTOPLASM
P
P
740
751
P
P
PI3K
GAP P
771
P
GAP
P
1009
P
PTP
PI3K
PTP
P
1021
PLCγ
P
PLCγ
phosphorylation sites. When expressed in cells that do not make a PDGF
receptor of their own, each of the receptors is phosphorylated at its tyrosines upon binding of PDGF. As shown in Figure 15–15, DNA synthesis is
stimulated to diferent extents in cells expressing the mutant receptors.
What roles do PI3K, GAP, PTP, and PLCγ play in the stimulation of
DNA synthesis by PDGF?
Figure 15–15 Stimulation of DNA
synthesis by the normal PDGF receptor
and by receptors missing some
phosphorylation sites (Problem 15–98).
Stimulation by the normal receptor is set
arbitrarily at 100%. The presence of a
phosphorylation site (P site) is indicated
by +; absence of a site by –.
15–99
50
protein
0
P site
1
2
3
4
5
6
7
8
9
PI3K
GAP
PTP
PLCγ
740, 751
771
1009
1021
+
+
+
+
+
–
–
–
–
+
–
–
–
–
+
–
–
–
–
+
+
–
–
+
–
+
–
+
–
–
+
+
–
–
–
–
MAP kinase kinase kinase (MAPKKK) activates MAP kinase kinase
(MAPKK) by phosphorylation of two serine side chains. Doubly phosphorylated (active) MAPKK, in turn, activates MAP kinase (MAPK) by
the phosphorylation of a threonine and a tyrosine. he doubly phosphorylated MAPK then phosphorylates a variety of target proteins to bring
about complex changes in cell behavior. It is possible to write down all of
the rate equations for the individual steps in this activation cascade, as
well as for the removal of the phosphates (inactivation) by protein phosphatase, and to solve them by making reasonable assumptions about the
concentrations of the proteins. he calculated plot of activation of the
kinases versus input stimulus is shown in Figure 15–16. Why is the very
steep response curve for MAPK a good thing for this signaling pathway?
15–100 An explicit assumption in the analysis in Problem 15–99 is that the
components of the MAP kinase module operate independently of one
another, so that the dual phosphorylation events that activate MAPKK
and MAPK occur one at a time as molecules collide in solution. How do
you suppose the curves in Figure 15–16 would change if a scafold protein
held the kinases of the MAP kinase cascade together? Most MAP kinase
modules are scafolded. What is the advantage of linking these kinases
onto scafold proteins?
MAPK
100
MAPKK
kinase activity
DNA synthesis
(% maximum)
100
MAPKKK
50
0
0
1
2
3
input stimulus
Figure 15–16 Stimulus–response curves
for the components of the MAPk cascade
(Problem 15–99). For ease of comparison,
the curves have been normalized so
that an input stimulus of 1 gives 50%
activation of the kinases.
324
Chapter 15: Cell Signaling
progesterone
Mos mRNA
(A) POOLED OOCYTES
– +
active (+ P )
Mos
% active
MAP kinase
inactive (– P)
100
MEK1
MAP kinase
50
0
0.001 0.01 0.1 1 10
progesterone (µM)
Figure 15–17 Progesterone-induced
MAP kinase activation, leading to
oocyte maturation (Problem 15–101).
MEk1 is the frog’s MAP kinase
kinase.
(B) INDIVIDUAL OOCYTES
mature
oocytes
–+
0.03 µM progesterone
0.1 µM progesterone
1 mm
0.3 µM progesterone
15–101 Activation (“maturation”) of frog oocytes is signaled through a MAP
kinase signaling module. An increase in the hormone progesterone triggers the module by stimulating the translation of the mRNA for Mos,
which is the frog’s MAP kinase kinase kinase (Figure 15–17). Maturation
is easy to score visually by the presence of a white spot in the middle of
the brown surface of the oocyte (Figure 15–17). To determine the dose–
response curve for progesterone-induced activation of MAP kinase,
you place 16 oocytes in each of six plastic dishes and add various concentrations of progesterone. After an overnight incubation, you crush
the oocytes, prepare an extract, and determine the state of MAP kinase
phosphorylation (hence, activation) by SDS polyacrylamide-gel electrophoresis (Figure 15–18A). his analysis shows a graded response of MAP
kinase to increasing concentrations of progesterone.
Before you crushed the oocytes, you noticed that not all oocytes in
individual dishes had white spots. Had some oocytes undergone partial
activation and not yet reached the white-spot stage? To answer this question, you repeat the experiment, but this time you analyze MAP kinase
activation in individual oocytes. You are surprised to ind that each
oocyte has either a fully activated or a completely inactive MAP kinase
(Figure 15–18B). How can an all-or-none response in individual oocytes
give rise to a graded response in the population?
15–102 Akt is a key protein kinase in the signaling pathway that leads to cell
growth. Akt is activated by a phosphatidylinositol-dependent protein kinase (PDK1), which phosphorylates threonine 308. At the same
time, serine 473 is phosphorylated. Your advisor has been unsuccessful in purifying the protein kinase responsible for the phosphorylation
of serine 473, but you think you know what is going on. You construct
genes encoding two mutant forms of Akt: one carries a point mutation
in the kinase domain, Akt-K179M, which renders it kinase-dead, and the
other carries a point mutation in the domain required to bind to PDK1
(Akt-T308A), which cannot be activated by PDK1. You transfect each of
these constructs, and a construct for wild-type Akt, into cells that do not
express their own Akt. You treat a portion of the cells with an insulin-like
growth factor (IGF1), which activates PDK1, and analyze the phosphorylation state of the various forms of Akt using antibodies speciic for Akt
or for particular phosphorylated amino acids (Figure 15–19).
What is the identity of the enzyme that phosphorylates serine 473 on
Akt?
Figure 15–18 Activation of frog oocytes
(Problem 15–101). (A) Phosphorylation
of MAP kinase in pooled oocytes.
(B) Phosphorylation of MAP kinase in
individual oocytes. MAP kinase was
detected by immunoblotting using a
MAP-kinase-speciic antibody. The
irst two lanes in each gel contain
nonphosphorylated, inactive Map
kinase (–) and phosphorylated, active
Map kinase (+).
construct
IGF1
Akt
Akt
T308A
Akt
K179M
–
+
–
+
–
+
1
2
3
4
5
6
anti-Akt
anti-P473
anti-P308
Figure 15–19 expression levels of
various forms of akt and their degree
of phosphorylation in the presence and
absence of IGF1 (Problem 15–102).
anti-akt recognizes all three forms of
akt regardless of their phosphorylation
state; anti-p473 speciically recognizes
the phosphorylated serine at position
473; anti-p308 speciically recognizes the
phosphorylated threonine at position 308.
ALTERNATIVE SIGNALING ROUTES IN GENE REGULATION
heptad
repeats
325
SH3 SH2 P
15–103 Interferon-γ (IFNγ) is a cytokine produced by activated T lymphocytes. It
binds to surface receptors on macrophages and stimulates their eicient
scavenging of invading viruses and bacteria via a JAK–STAT signaling
pathway. A number of genes are activated in response to IFNγ binding,
all of which contain a DNA sequence element with partial dyad symmetry (TTCCXGTAA) that is required for the IFNγ response.
You have cloned the gene for the STAT transcription factor that is
activated in response to IFNγ binding. he sequence of the gene indicates that the protein contains several heptad repeat sequences near its
N-terminus—a common dimerization domain in many transcription
factors—and SH2 and SH3 domains adjacent to a site for tyrosine phosphorylation near the C-terminus (Figure 15–20). By making antibodies
to the protein, you show that it is normally located in the cytosol. After
15 minutes exposure to IFNγ, the protein becomes phosphorylated on a
tyrosine and moves to the nucleus.
Suspecting that tyrosine phosphorylation is the key to the regulation
of this transcription factor, you assay its ability to bind the DNA sequence
element in the presence of high concentrations of free phosphotyrosine
or when mixed with anti-phosphotyrosine antibodies. Both treatments
inhibit binding of the protein to DNA, as does treatment with a protein
phosphatase. Finally, you measure the molecular weight of the cytosolic
and nuclear forms of the protein, which suggest that the cytosolic form is
a monomer and the nuclear form is a dimer.
A. Do you think that phosphorylation of the transcription factor is necessary for the factor to bind to DNA, or do you think phosphorylation is
required to create an acidic activation domain to promote transcription?
B. Bearing in mind that SH2 domains bind phosphotyrosine, how do you
suppose free phosphotyrosine might interfere with the activity of the
transcription factor?
C. How might tyrosine phosphorylation of the protein promote its dimerization? How do you think dimerization enhances its binding to DNA?
ALTERNATIVE SIGNALING ROUTES IN GENE
REGULATION
TERMS TO LEARN
β-catenin
circadian clock
Cubitus interruptus (Ci)
Delta
Dishevelled
Frizzled
Hedgehog protein
iHog
IκB
LDL-receptor-related
protein (LRP)
NFκB proteins
Notch
nuclear receptor
superfamily
Patched
Smoothened
steroid hormone
Wnt/β-catenin
pathway
Wnt proteins
DEFINITIONS
Match each deinition below with its term from the list above.
15–104 Receptor protein involved in what may be the most widely used signal-
ing pathway in animal development; its ligands are cell-surface proteins
such as Delta.
Figure 15–20 Sequence elements in the
transcription factor that responds to IFNγ
(Problem 15–103).
326
Chapter 15: Cell Signaling
15–105 A family of secreted signal molecules that act as local mediators and
morphogens during development; they were initially discovered as the
products of the Wingless gene in lies and the Int1 gene in mice.
15–106 A signaling pathway activated by Wnt binding to both the Frizzled recep-
tor and the LRP co-receptor.
15–107 A group of secreted signal molecules that act as local mediators and mor-
phogens during development and whose efects are mediated through
the cell-surface receptor Patched and its binding partner Smoothened.
15–108 A target of Hedgehog signaling, this gene regulatory molecule is a full-
length gene activator in the presence of Hedgehog and a partially proteolyzed gene repressor in its absence.
15–109 Latent gene regulatory proteins that are present in most cells in both ani-
mals and plants and are central to many stress, inlammatory, and innate
immune responses.
15–110 Hydrophobic signaling molecule with a characteristic four-ringed struc-
ture derived from cholesterol.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
15–111 Signaling pathways that activate latent gene regulatory proteins depend
on regulated proteolysis to control activity and location.
15–112 Notch is both a cell-surface receptor and a latent gene regulatory protein.
15–113 Because one of the targets of NFκB activation is the gene for IκBα, the
cytoplasmic inhibitor of NFκB, a negative feedback loop is established
that limits the duration of the NFκB response.
THOUGHT PROBLEMS
15–114 Why do signaling responses that involve changes in proteins already
present in the cell occur in milliseconds to seconds, whereas responses
that require changes in gene expression require minutes to hours?
15–115 Like Notch, the β-amyloid precursor protein (APP) is cleaved near its
transmembrane segment to release an extracellular and an intracellular component. Explain how the fragments of APP relate to the amyloid
plaques that are characteristic of Alzheimer’s disease.
15–116 he Wnt planar polarity signaling pathway normally ensures that each
wing cell in Drosophila has a single hair. Overexpression of the Frizzled
gene from a heat-shock promoter (hs-Fz) causes multiple hairs to grow
from many cells (Figure 15–21A). his phenotype is suppressed if hs-Fz is
combined with a heterozygous deletion (DshΔ/+) of the Dishevelled gene
(Figure 15–21B). Do these results allow you to order the action of Frizzled and Dishevelled in the signaling pathway? If so, what is the order?
Explain your reasoning.
(A)
(B)
15–117 here are two common mutational routes to the uncontrolled cell pro-
liferation and invasiveness that characterize cancer cells. he irst is to
make a stimulatory gene (a proto-oncogene) hyperactive: this type of
mutation has a dominant efect so that only one of the cell’s two gene
copies needs to undergo change. he second is to make an inhibitory
gene (a tumor suppressor gene) inactive: this type of mutation usually is
recessive so that both the cell’s gene copies must be inactivated.
Mutations of the Apc (adenomatous polyposis coli) gene occur in 80%
hs-Fz / +
+/+
hs-Fz / +
Dsh ∆ / +
Figure 15–21 Pattern of hair growth
on wing cells in genetically different
Drosophila (Problem 15–116).
ALTERNATIVE SIGNALING ROUTES IN GENE REGULATION
of human colon cancers. Normal APC increases the ainity of the degradation complex for β-catenin, which in excess can enter the nucleus and
promote transcription of key target genes for cell proliferation. Given this
information, which category—oncogene or tumor suppressor—would
you expect the Apc gene to belong to? Why?
327
HO
cholesterol
15–118 Latent gene regulatory proteins are prevented from entering the nucleus
until the cell receives an appropriate signal. List four ways by which cells
keep gene regulatory proteins out of the nucleus, and give an example of
a latent gene regulatory protein that is controlled by each mechanism.
CH2OH
C O
OH
HO
15–119 he steroid hormones cortisol, estradiol, and testosterone are all derived
from cholesterol by modiications that introduce polar groups such as –
OH and =O (Figure 15–22). If cholesterol itself was not normally found
in cell membranes, do you suppose it could be used efectively as a hormone, provided that an appropriate intracellular receptor was available?
O
cortisol
15–120 Most people who are completely blind have circadian rhythms that
are ‘free-running;’ that is, their circadian rhythms are not synchronized to environmental time cues and they oscillate on a cycle of about
24.5 hours. Why do you suppose the circadian clocks of blind people are
not entrained to the same 24-hour clock as the majority of the population? Can you guess what symptoms might be associated with a freerunning circadian clock? Do you suppose that blind people have trouble
sleeping?
OH
O
testosterone
OH
DATA HANDLING
15–121
β-Catenin can be phosphorylated by glycogen synthase kinase 3 (GSK3)
and it can be degraded in proteasomes. β-Catenin could be sensitized for
degradation by phosphorylation, it could be protected from degradation
by phosphorylation, or its phosphorylation status could be irrelevant for
degradation. To distinguish among these possibilities, you generate cell
lines that express either a mutant GSK3 that cannot carry out phosphorylation, or a mutant β-catenin that is missing its site of phosphorylation.
In the presence and absence of the proteasome inhibitor, ALLN, both
cell lines yield β-catenin that migrates as a single band, with no slower
migrating bands visible. In contrast, nonmutant β-catenin and GSK3
display several slower migrating bands in the presence of ALLN, but no
slower migrating bands in its absence. What is the relationship between
β-catenin phosphorylation and its degradation in proteasomes? Explain
your answer.
HO
estradiol
Figure 15–22 Steroid hormones and
their parent molecule, cholesterol
(Problem 15–119).
(A) HEDGEHOG PRECURSOR
G257
C258
1
N-
471
-C
(B) TIME COURSE
15–122 he Hedgehog gene encodes the Hedgehog precursor protein, which is
471 amino acids long. he precursor protein (Figure 15–23A) is normally
cleaved between glycine 257 (G257) and cysteine 258 (C258) to generate a fragment that is active in local and long-range signaling. Cleavage
is essential for signaling. You clone a segment of the Hedgehog gene
encoding a portion of the protein that includes the cleavage site and
the entire C-terminus. When you purify this protein and incubate it in
bufer, you observe cleavage over the course of several hours, as shown in
Figure 15–23B. If you vary its concentration over a 256-fold range and
assay cleavage after 4 hours of incubation, you observe the results shown
in Figure 15–23C.
A. Explain how these data support the idea that the Hedgehog precursor
protein cleaves itself. How do they rule out the possibility that the puriied protein is contaminated with a bacterial protease, for example?
B. Does a molecule of precursor protein cleave itself, or does it cleave
another molecule of the precursor; that is, is the reaction intramolecular
or intermolecular?
0
0.5 1 2 4 8 16
hours
(C) CONCENTRATION DEPENDENCE
0.05 0.2 0.8 3.2 12.8
µM
Figure 15–23 Mechanism of cleavage of
the Hedgehog precursor protein (Problem
15–122). (A) Site of cleavage (red arrow)
in the Hedgehog precursor protein.
(B) Time course of cleavage of the
fragment of the precursor protein.
(C) Dependence of cleavage on
concentration of the precursor protein
fragment.
328
Chapter 15: Cell Signaling
(C) CELLS
wt
1–257 (N)
(B) FLIES
iu
m
m
ed
m
iu
lls
ed
ce
lls
m
G257
ce
ve
ct
w or
t
1–
25
un 7 (
cl N )
ea
va
bl
e
(A) CONSTRUCTS
5
6
7
8
C258
471
1
wild type
1
A258
Figure 15–24 Fate of the fragments of
Hedgehog after cleavage (Problem 15–
123). (A) Constructs encoding different
forms of the Hedgehog precursor protein.
(B) Results of expression in Drosophila
embryos. (C) Results of expression in
insect cells. Hedgehog fragments were
detected using antibodies speciic for the
n-terminal segment.
471
uncleavable
1
257
1–257 (N)
N
1
2
3
4
15–123 To ind out what happens to the fragments of Hedgehog after cleavage,
you express three versions: wild-type Hedgehog precursor, an uncleavable form, and the N-terminal cleavage product (Figure 15–24A). In
ly embryos, the constructs behave as expected: wild-type Hedgehog is
cleaved, the uncleavable version is not, and the N-terminal segment is
expressed (Figure 15–24B). When wild-type Hedgehog and the N-terminal segment are expressed in insect cells, however, the N-terminal segment from wild-type Hedgehog remains associated with the cells, while
the synthesized N-terminal segment is secreted into the medium (Figure 15–24C). Can you suggest possible explanations for the diference in
localization of the N-terminal segment?
15–124 If you overexpress various Hedgehog constructs (see Figure 15–24A) in
normal ly embyros and examine the pattern of Wnt expression (a wellcharacterized target of Hedgehog signaling), you observe a striped pattern of expression in all cases, but some constructs lead to thicker stripes
than normal (Figure 15–25).
A. Which part of the Hedgehog molecule is responsible for signaling?
B. All the cells in the embryo are overexpressing the various Hedgehog
constructs. Why is it, do you suppose, that you observe the same basic
striped pattern of Wnt expression in all of them?
C. Why do you see stripes of Wnt expression even in the absence of Hedgehog overexpression?
15–125 Studies with the fruit ly Drosophila provided initial clues to the complex
changes in patterns of gene expression that a simple hormone can trigger. Drosophila larvae molt in response to an increase in the concentration of the steroid hormone ecdysone. he polytene chromosomes of the
Drosophila salivary glands are an excellent experimental system in which
to study the pattern of gene activity initiated by the hormone because
active genes enlarge into pufs that are visible in the light microscope.
Furthermore, the size of a puf is proportional to the rate at which the
gene is transcribed. Prior to addition of ecdysone, a few pufs—termed
intermolt pufs—are already active. Upon exposure of dissected salivary
glands to ecdysone, these intermolt pufs regress, and two additional sets
forms of Hedgehog
vector
1
wild type
2
uncleavable
3
1–257 (N)
4
257–471 (C)
5
Figure 15–25 patterns of Wnt
expression in Drosophila embryos that
are overexpressing various hedgehog
constructs (Problem 15–124). Wnt
expression was detected by in situ
hybridization.
SIGNALING IN PLANTS
329
puff size (% maximum)
(A) NORMAL
intermolt
100
early
late
50
0
0
2
4
6
8
10
12
10
12
10
12
time (hours)
add
ecdysone
(B) CYCLOHEXIMIDE
puff size (% maximum)
of pufs appear. he early pufs arise within a few minutes after addition
of ecdysone; the late pufs arise within 4–10 hours. he concentration of
ecdysone does not change during this time period. he pattern of puf
appearance and disappearance is illustrated for a typical puf in each category in Figure 15–26A.
Two critical experiments helped to deine the relationships between
the diferent classes of puf. In the irst, cycloheximide, which blocks protein synthesis, was added at the same time as ecdysone. As illustrated
in Figure 15–26B, under these conditions the early pufs did not regress
and the late pufs were not induced. In the second experiment, ecdysone
was washed out after a 2-hour exposure. As illustrated in Figure 15–26C,
this treatment caused an immediate regression of the early pufs and a
premature induction of the late pufs.
A. Why do you think the early pufs didn’t regress and the late pufs weren’t
induced in the presence of cycloheximide? Why do you think the intermolt pufs were unafected?
B. Why do you think the early pufs regressed immediately when ecdysone
was removed? Why do you think the late pufs arose prematurely under
these conditions?
C. Outline a model for ecdysone-mediated regulation of the puing pattern.
100
intermolt
early
2
6
50
0
0
4
8
time (hours)
add
ecdysone
SIGNALING IN PLANTS
(C) ECDYSONE PULSE
leucine-rich repeat (LRR) receptor kinase
phototropin
phytochrome
plant growth regulator (plant hormone)
DEFINITIONS
Match each deinition below with its term from the list above.
15–126 A cytoplasmic serine/threonine kinase in plants that is activated by red
light and inactivated by far-red light.
15–127 Small gas molecule inluential in various aspects of plant development,
including fruit ripening and leaf abscission.
15–128 General term for a signal molecule that helps coordinate growth and
development in plants.
15–129 Flavoprotein responsive to blue light, found in both plants and animals;
in animals it is involved in circadian rhythms.
15–130 A growth regulator that helps plants grow toward light, grow upward
rather than branch out, and extend their roots downward.
15–131 Common type of receptor serine/threonine kinase in plants, character-
ized by an extracellular portion rich in repeated segments containing a
high proportion of leucine.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
15–132 Even though plants and animals independently evolved multicellularity,
they use virtually all the same signaling proteins and second messengers
for cell–cell communication.
15–133 Remarkably, the auxin elux transporters in the cap cells of the root
quickly redistribute themselves in response to a change in the direction
puff size (% maximum)
TERMS TO LEARN
auxin
brassinosteroids
cryptochrome
ethylene
intermolt
100
late
early
50
0
0
2
add
ecdysone
4
6
8
time (hours)
remove
ecdysone
Figure 15–26 Pufing patterns in
Drosophila salivary gland giant
chromosomes (Problem 15–125).
(a) normal pufing pattern.
(B) pufing pattern in the presence
of cycloheximide. (C) pufing pattern
after removal of ecdysone.
330
Chapter 15: Cell Signaling
(A)
(B)
(C)
plants fungi animals
plants fungi animals
plants animals fungi
of the gravity vector, so that they pump auxin toward the side of the root
pointing downward.
THOUGHT PROBLEMS
15–134 he last common ancestor to plants and animals was a unicellular eukar-
yote. hus, it is thought that multicellularity and the attendant demands
for cell communication arose independently in these two lineages. his
evolutionary viewpoint accounts nicely for the vastly diferent mechanisms that plants and animals use for cell communication. Fungi use
signaling mechanisms and components that are very similar to those
used in animals. Which of the phylogenetic trees shown in Figure 15–27
does this observation support?
15–135 If signaling arose as a solution to the demands of multicellularity, how
then do you account for the very similar mechanisms of signaling that are
used in animals and the unicellular fungus Saccharomyces cerevisiae?
15–136 How is it that plant growth regulators can be present throughout a plant
and yet have speciic efects on particular cells and tissues?
DATA HANDLING
15–137 he ripening of fruit is a complicated process of development, difer-
entiation, and death (except for the seeds, of course). he process is
triggered by minute amounts of ethylene gas. (his was discovered by
accident many years ago; the parain stoves used to heat greenhouses
in the olden days gave of enough ethylene to initiate the process.) he
ethylene is normally produced by the fruits themselves in a biochemical
pathway, the rate-limiting step of which is controlled by ACC synthase,
which converts S-adenosylmethionine to a cyclopropane compound
that is the immediate precursor of ethylene. Ethylene initiates a program
of sequential gene expression that includes the production of several
new enzymes, including polygalacturonase, which probably contributes
to softening the cell wall.
Your company, Agribucks, is trying to make mutant tomatoes that
cannot synthesize their own ethylene. Such fruit could be allowed to stay
longer on the vine, developing their lavor while remaining green and
irm. hey could be shipped in this robust unripe state and exposed to
ethylene just before arrival at market. his should allow them to be sold
at the peak of perfection, and the procedure involves no artiicial additives of any kind.
You decide to use an antisense approach, which works especially well
in plants. You place an ACC synthase cDNA into a plant expression vector
so that the gene will be transcribed in reverse, introduce it into tomato
cells, and regenerate whole tomato plants. Sure enough, ethylene production is inhibited by 99.5% in these transgenic tomato plants, and their
fruit fails to ripen. But when placed in air containing a small amount of
ethylene, they turn into beautiful, tasty, ripe red fruit in about 2 weeks.
A. How do you imagine that transcribing the ACC synthase gene in reverse
blocks the production of ethylene?
B. Will you be a millionaire before you are 30?
Figure 15–27 Three possible phylogenetic
relationships among plants, animals, and
fungi (Problem 15–134).
SIGNALING IN PLANTS
MCAT STYLE
Passage 1 (Questions 15–138 to 15–140)
he Ras GTPase was irst discovered as a gene that plays an important role in
transforming normal cells into cancer cells. Although Ras is normally activated by
a receptor tyrosine kinase (RTK), in many kinds of cancer the Ras gene has sustained a mutation that makes the Ras protein hyperactive. his mutant Ras sends
unregulated signals that drive cell proliferation and contribute to tumor formation. Activated Ras binds and activates a MAP kinase kinase kinase (MAPKKK)
called Raf, which activates a MAP kinase kinase (MAPKK) called Mek, which then
activates a MAP kinase (MAPK) called Erk.
15–138 What kinds of mutations in the Ras gene could lead to hyperactive Ras?
I. Mutations that stimulate Ras to bind the Ras GTPase-activating protein
II. Mutations that decrease the ability of Ras to hydrolyze GTP
III. Mutations that block Ras binding to Ras-GEF
A. I
B. II
C. I and III
D. II and III
15–139 Mutant forms of Raf have also been found to play an important role in
cancer. A mutant called Raf-V600E causes Raf to become hyperactive
independently of signals from Ras. Drugs that inhibit Raf-V600E cause
rapid regression of tumors that express Raf-V600E. It was recently discovered that treatment of cancer cells with these drugs increases Ras activity. Which of the following hypotheses best explains this observation?
A. Erk normally phosphorylates and inhibits Raf to restrict the duration of
RTK signaling. Inhibition of Erk therefore leads to increased Ras activity.
B. Erk normally phosphorylates the RTK and inhibits its signaling. Inhibition of Raf-V600E decreases Erk activity, which leads to increased RTK
signaling.
C. Erk normally phosphorylates the RTK and stimulates its signaling. Inhibition of Raf-V600E increases RTK signaling, which leads to increased
Ras activity.
D. Raf normally phosphorylates the RTK and stimulates its signaling. Inhibition of Raf-V600E therefore increases RTK signaling and increases Ras
activity.
15–140 Imagine you are working in a cancer clinic and encounter a patient with
a cancer that has the Raf-V600E mutation. You treat with a Raf inhibitor, but the cancer does not respond. You are working on developing a
new treatment plan. Which of the following drugs would make the most
sense?
A. An inhibitor of Ras-GEF
B. An inhibitor of Erk
C. An inhibitor of Ras
D. An inhibitor of the RTK
Passage 2 (Questions 15–141 to 15–143)
Scafold proteins are thought to constrain signaling speciicity by bringing multiple kinases into close proximity to ensure that they signal to each other, rather
than to other proteins in the cell. he role of scafolds in signaling was elucidated
in studies aimed at understanding the speciicity of MAP kinase cascades in yeast.
In the MAP kinase cascade that prepares the cell for mating, an extracellular mating pheromone activates a G protein, which then activates a MAP kinase cascade
that includes Ste11 (MAPKKK), Ste7 (MAPKK), and Fus3 (MAPK) (Figure 15–28).
Activation of this cascade occurs over a time scale of 5–10 minutes. he MAP
kinase cascade that controls the response to starvation is activated by the Ras
331
Chapter 15: Cell Signaling
GTPase and also includes Ste11 and Ste7; however, the pathway works through a
MAPK called Kss1, rather than Fus3. his cascade is activated over a time scale of
several hours. How can activation of the same kinases—Ste11 and Ste7—lead to
completely diferent outputs? he discovery that Ste5 binds to the G protein and
to all of the MAP kinases in the mating-response pathway led to the idea that Ste5
acts as a scafold to sequester the MAP kinases and link their activation to the activation of the G protein. Recent work suggests that scafolds may play even more
complex and interesting roles.
15–141 Which of the following observations would make you question the
sequestration model for scafolds?
A. Ste7 activates Fus3 and Kss1 in vitro in the absence of Ste5.
B. Ste7 dissociates from Ste5 with a half-life of 5 seconds.
C. he Ste5 scafold binds to Fus3, but does not bind to Kss1.
D. When activated in the mating pathway, Fus3 inactivates Kss1.
15–142 In one series of experiments, puriied proteins were used to measure the
ability of Ste7 to phosphorylate Fus3 and Kss1. Kinase reactions were carried out in the presence or absence of a domain of Ste5 that was found
to play an important role in Fus3 activation. Ste7 robustly phosphorylated Kss1 by itself, and addition of the Ste5 domain had no efect on the
Km or Kcat of the reaction. In contrast, the Ste5 domain gave a 5000-fold
increase in the rate of phosphorylation of Fus3 by Ste7, with little efect
on the Km. Which one of the following hypotheses could explain these
experimental results?
A. Fus3 induces a conformational change in Ste5 that activates Ste7.
B. Ste5 alters the conformation of Fus3 to allow phosphorylation by Ste7.
C. he binding of the Ste5 domain to Ste7 activates its kinase activity.
D. he Ste5 domain positions Ste7 so that it binds more tightly to Fus3.
15–143 In another series of in vitro experiments, activation of Fus3 by Ste7 was
measured in the presence of either full-length Ste5 or the Ste5 domain
that activates Fus3. he rate of activation of Fus3 in the presence of fulllength Ste5 was 10-fold lower than in the presence of the Ste5 domain. In
addition, it was found that cells expressing the Ste5 domain, instead of
full-length Ste5, inappropriately activated Fus3 in response to starvation.
Which of the following hypotheses would explain these observations?
I. Full-length Ste5 inhibits Fus3 activation in the absence of mating pheromone, ensuring that starvation signals relayed by Ste7 cannot activate
Fus3.
II. Full-length Ste5 contains a domain that promotes feedback activation of
Ste7 by Fus3.
III. In the absence of mating pheromone, full-length Ste5 is in a conformation that inhibits its ability to facilitate Fus3 activation. Mating-pheromone signaling triggers a conformational change in Ste5 that relieves the
inhibition.
A. I
B. II
C. I and III
D. II and III
mating
Ste5
332
starvation
Ste11
MAPKKK
Ste11
Ste7
MAPKK
Ste7
Fus3
MAPK
Kss1
Figure 15–28 Mating-pheromone
activation of the MAP kinase cascade in
yeast (Problems 15–141 to 15–143).
Chapter 16
333
CHAPTER
The Cytoskeleton
16
FUNCTION AND ORIGIN OF THE CYTOSKELETON
IN THIS CHAPTER
TERMS TO LEARN
cytoskeleton
motor protein
protofilament
DEFINITIONS
Match each deinition below with its term from the list above.
16–1
16–2
A linear chain of protein subunits joined end to end, which associates
laterally with other such chains to form cytoskeletal components.
System of protein ilaments in the cytoplasm of a eukaryotic cell that
gives the cell its shape and the capacity for directed movement.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
16–3
Microtubules determine the shape of the cell’s surface and are necessary
for whole-cell locomotion, and drive the pinching of one cell into two.
16–4
Even though the actin bundles at the cores of stereocilia on the hair cells
of the inner ear maintain their stable organization for the entire lifetime
of the animal, they are continuously remodeled and replaced on average
every 48 hours.
16–5
Because bacteria lack the elaborate networks of intracellular membrane-enclosed organelles typical of eukaryotic cells, they do not require
cytoskeletal ilaments.
THOUGHT PROBLEMS
16–6
In general terms, what are the cellular functions of intermediate ilaments, microtubules, and actin ilaments?
16–7
If each type of cytoskeletal ilament is made up of subunits that are held
together by weak noncovalent bonds, how is it possible for a human
being to lift heavy objects?
16–8
List diferences between bacteria and animal cells that could have
depended on the appearance during evolution of some or all of the components of the present eukaryotic cytoskeleton. Why do you suppose
a cytoskeleton might have been crucial for each of these diferences to
evolve?
16–9
he amino acid sequences of actins and tubulins from all eukaryotes are
remarkably well conserved, yet the large numbers of proteins that interact with these ilaments are no more conserved than most other proteins
FUNCTION AND ORIGIN OF
THE CYTOSKELETON
ACTIN AND ACTIN-BINDING
PROTEINS
MYOSIN AND ACTIN
MICROTUBULES
INTERMEDIATE FILAMENTS
AND SEPTINS
CELL POLARIZATION AND
MIGRATION
Chapter 16: The Cytoskeleton
334
in diferent species. How can it be that the ilament proteins themselves
are highly conserved, while the proteins that interact with them are not?
CALCULATIONS
16–10
he average time it takes particles to difuse a distance of x cm is
t = x2/2D
where t is the time in seconds and D is the difusion coeicient, which is
a constant that depends on the size and shape of the particle.
A. How long would it take for a small molecule, a protein molecule, and a
membrane-enclosed vesicle to difuse across a cell 10 μm in diameter? A
typical difusion coeicient for a small molecule is 5 × 10–6 cm2/sec, for
a protein molecule 5 × 10–7 cm2/sec, and for a membrane vesicle 5 × 10–8
cm2/sec.
B. Why do you suppose a cell relies on the strategy of polymerizing and
depolymerizing cytoskeletal ilaments, rather than on difusion of the
ilaments themselves, to accomplish its cytoskeletal rearrangements?
DATA HANDLING
One of the most striking examples of a purely actin-based cellular movement is the extension of the acrosomal process of a sea cucumber sperm.
he sperm contains a store of unpolymerized actin in its head. When a
sperm makes contact with a sea cucumber egg, the actin polymerizes
rapidly to form a long spearlike extension. he tip of the acrosomal process penetrates the egg, and it is probably used to pull the sperm inside.
Are actin monomers added to the base or to the tip of the acrosomal
bundle of actin ilaments during extension of the acrosomal process? If
the supply of monomers to the site of assembly depends on difusion, it
should be possible to distinguish between these alternatives by measuring the length of the acrosomal process with increasing time. If actin
monomers are added to the base of the process, which is inside the head,
the rate of growth should be linear because the distance between the site
of assembly and the pool of monomers does not change with time. On the
other hand, if the subunits are added to the tip, the rate of growth should
decline progressively as the acrosomal process gets longer because the
monomers must difuse all the way down the shaft of the process. In this
case, the rate of extension should be proportional to the square root of
time. Plots of the length of the acrosomal process versus time and the
square root of time are shown in Figure 16–1.
A. Are the ascending portions of the plots in Figure 16–1 more consistent
with the addition of actin monomers to the base or to the tip of the acrosomal process?
B. Why do you suppose the process grows so slowly at the beginning and at
the end of the acrosomal reaction?
16–11
length ( µm)
80
60
40
20
0
0
2
4
6
time (seconds)
8
10
0
1
2
3
4
square root of time ( seconds0.5 )
Figure 16–1 Plots of the length of the
acrosome versus time and the square root
of time (Problem 16–11).
ACTIN AND ACTIN-BINDING PROTEINS
335
ACTIN AND ACTIN-BINDING PROTEINS
fluorescence intensity
TERMS TO LEARN
Arp2/3 complex
cell cortex
4
formin
treadmilling
DEFINITIONS
Match each deinition below with its term from the list above.
16–12
he process by which a polymeric protein ilament is maintained at constant length by addition of protein subunits at one end and loss of subunits at the other.
16–13
Specialized layer of cytoplasm on the inner face of the plasma membrane, rich in actin ilaments.
16–14
Protein assembly that nucleates actin ilament growth from the minus
end, allowing rapid growth at the plus end and forming a treelike web of
ilaments.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
16–15
In the treelike web of actin ilaments that form the cell cortex, an Arp2/3
complex anchors each actin ilament branch to the side of another actin
ilament.
16–16
All the proteins that bind to the ends of actin ilaments cap the ends to
prevent further polymerization.
C
3
B
2
A
1
0
0
30
60
90
time (seconds)
Figure 16–2 Formation of actin ilaments
over time, starting with puriied actin
monomers that are labeled with a
luorescent probe (Problem 16–17).
Upon polymerization, the luorescence
of the probe increases, which allows
polymerization to be measured. The
intensity of luorescence at zero seconds
is due to the background luorescence of
the actin monomers. The three phases of
polymerization are indicated as A, B, and
C. Fluorescence intensity is measured in
arbitrary units.
THOUGHT PROBLEMS
A typical time course of polymerization of actin ilaments from actin subunits is shown in Figure 16–2.
A. Explain the properties of actin polymerization that account for each of
the three phases of the polymerization curve.
B. How would the curve change if you doubled the concentration of actin?
Would the concentration of free actin at equilibrium be higher or lower
than in the original experiment, or would it be the same in both?
16–17
16–18
Figure 16–3 shows the equilibrium distribution of actin in free subunits
(monomers) and in ilaments, as a function of actin concentration. Indicate the critical concentration of actin on this diagram.
16–19
Imagine that the polymer in Figure 16–4A can add subunits at either
end, just like actin ilaments (and microtubules). Imagine also three
hypothetical types of free subunit, as shown in Figure 16–4B. Each type
(A) POLYMERIZATION
mass
filament
monomer
or
actin concentration
Figure 16–3 Mass of actin monomers and
ilaments as a function of actin concentration
(Problem 16–18).
(B) SUBUNIT CONFORMATIONS
1
2
3
Figure 16–4 Polymerization of a polymer
(Problem 16–19). (A) Addition of a subunit
to a polymer. (B) Conformations of three
hypothetical subunits.
Chapter 16: The Cytoskeleton
336
of subunit can add to the polymer and, once added, it adopts the conformation of the other subunits in the polymer (Figure 16–4A). For each of
these subunits, decide which end of the polymer, if either, will grow at the
faster rate when the concentration of that subunit is higher than the critical concentration required for polymerization. Explain your reasoning.
For any of the subunits, will there be a concentration at which one end
will preferentially grow while the other shrinks? Why or why not?
16–20
Some actin-binding proteins signiicantly increase the rate at which the
formation of actin ilaments is initiated in the cytosol. How might such
proteins do this? What must they not do when binding the actin monomers?
16–21
he concentration of actin in cells is 50–100 times greater than the critical
concentration observed for pure actin in a test tube. How is this possible?
What prevents the actin subunits in cells from polymerizing into ilaments? Why is it advantageous to the cell to maintain such a large pool of
actin subunits?
16–22
Coilin preferentially binds to older actin ilaments and promotes their
disassembly. How does coilin distinguish old ilaments from new ones?
Figure 16–5 Myosin-decorated actin
ilament after a few minutes in a solution
with excess actin monomers (Problem
16–23). The shorter, thicker segment is
the myosin-decorated actin ilament.
DATA HANDLING
If you add short actin ilaments marked by bound myosin heads (myosindecorated ilaments) to a solution with an excess of actin monomers, wait
for a few minutes, and then examine the ilaments by electron microscopy, you see the picture shown in Figure 16–5.
A. Which is the plus end of the myosin-decorated ilament and which is the
minus end? Which is the “barbed” end and which is the “pointed” end?
How can you tell?
B. If you diluted the mixture so that the actin concentration was below the
critical concentration, which end would depolymerize more rapidly?
C. When the actin ilament depolymerizes, why are subunits removed
exclusively from the ends and not from the middle of the ilament?
16–23
growth rate
(molecules per second)
200
plus end
100
minus end
0
0
5
10
15
actin (μM)
15
plus end
10
5
AB
C
D
minus end
0
E
Figure 16–6 Growth rates at the plus and minus ends of actin ilaments
as a function of actin concentration (Problem 16–24). (A) Measurements of
growth rates over a broad range of actin concentrations. (B) Growth rates
at low actin concentrations, shown on an expanded scale.
20
(B) EXPANDED SCALE
growth rate
(molecules per second)
he growth rates at the plus and minus ends of actin ilaments as a
function of actin concentration are shown in Figure 16–6A and, on an
expanded scale, in Figure 16–6B.
A. he data in Figure 16–6A were gathered by measuring initial growth rates
at each actin concentration. Similar data gathered for any Michaelis–
Menten enzyme would generate a hyperbolic plot, instead of the linear
plots shown here. Why does the growth rate of actin ilaments continue
to increase linearly with increasing actin concentration, whereas an
enzyme-catalyzed reaction reaches a plateau with increasing substrate
concentration?
B. Figure 16–6B shows the ilament growth rates at low actin concentration
on an expanded scale. Imagine that you could add actin ilaments to a
solution of actin subunits at the concentrations indicated as A, B, C, D,
and E. For each of these concentrations, decide whether the added actin
ilament would grow or shrink at its plus and minus ends. What is the
critical concentration for the plus end? What is the critical concentration
for the minus end? Would treadmilling occur at any of these concentrations?
16–24
(A) MEASUREMENTS
300
–5
0
0.5
actin (μM)
1.0
337
light scattering
100
20
ATP hydrolysis
10
0
0
100
200
50
Figure 16–7 The kinetics of actin
polymerization and ATP hydrolysis
(Problem 16–25).
0
300
time (seconds)
Your ultimate goal is to understand human consciousness, but your advisor wants you to understand some basic facts about actin assembly irst.
He tells you that ATP binds to actin monomers and is required for assembly. But, ATP hydrolysis is not necessary for polymerization since ADP
can, under certain circumstances, substitute for the ATP requirement.
ADP ilaments, however, are much less stable than ATP ilaments, supporting your secret suspicion that the free energy of ATP hydrolysis really
is used to drive actin assembly.
Your advisor suggests that you make careful measurements of the
quantitative relationship between the number of ATP molecules hydrolyzed and the number of actin monomers linked into polymer. he
experiments are straightforward. To measure ATP hydrolysis, you add
γ32P-ATP to a solution of polymerizing actin, take samples at intervals,
and determine how much radioactive phosphate has been produced. To
assay polymerization, you measure the increase in light scattering that
is caused by formation of the actin ilaments. Your results are shown in
Figure 16–7. Your light-scattering measurements indicate that 20 μmoles
of actin monomers were polymerized. Since the number of polymerized
actin monomers matches exactly the number of ATP molecules hydrolyzed, you conclude that one ATP is hydrolyzed as each new monomer is
added to an actin ilament.
When you show your advisor the data and tell him your conclusions,
he smiles and very gently tells you to look more closely at the graph. He
says your data prove that actin can polymerize without ATP hydrolysis.
A. What does your advisor see in the data that you have overlooked?
B. What do your data imply about the distribution of ATP and ADP in
polymerizing actin ilaments?
– cytochalasin B
0.8
viscosity
150
30
light scattering
(arbitrary units)
ATP hydrolyzed (μmol)
ACTIN AND ACTIN-BINDING PROTEINS
0.6
+ cytochalasin B
0.4
0.2
0.0
0
5
10
15
time (minutes)
20
16–25
Cytochalasin B strongly inhibits certain forms of cell motility, such
as cytokinesis and the ruling of growth cones, and it dramatically
decreases the viscosity of gels formed with mixtures of actin and a wide
variety of actin-binding proteins. hese observations suggest that cytochalasin B interferes with the assembly of actin ilaments. In the classic
experiment that deined its mechanism, short lengths of actin ilaments
were decorated with myosin heads and then mixed with actin subunits
in the presence or absence of cytochalasin B. Assembly of actin ilaments
was measured by assaying the viscosity of the solution (Figure 16–8) and
by examining samples by electron microscopy (Figure 16–9).
A. Suggest a plausible mechanism to explain how cytochalasin B inhibits
actin ilament assembly. Account for the appearance of the ilaments
in the electron micrographs and the viscosity measurements (both the
altered rate and extent).
B. he normal growth characteristics of an actin ilament and the actinbinding properties of cytochalasin B argue that actin monomers undergo
a conformational change upon addition to an actin ilament. How so?
Figure 16–8 Increase in the viscosity
of actin solutions in the presence and
absence of cytochalasin B (Problem
16–26).
– cytochalasin B
minus
end
plus
end
16–26
16–27
Phalloidin, which is a toxic peptide from the mushroom Amanita phalloides, binds to actin ilaments. Phalloidin tagged with a luorescent
probe is commonly used to stain actin ilament assemblies in cells
+ cytochalasin B
Figure 16–9 Appearance of typical actin
ilaments formed in the presence and
absence of cytochalasin B (Problem
16–26). The decorated actin ilaments
present before the addition of actin
monomers are shown at the top of each
set of three. Filaments present after
increasing times of incubation with actin
monomers (red circles) are shown below.
Chapter 16: The Cytoskeleton
338
(A) CELLS
(B) LOW
(C) HIGH
11 nm
Figure 16–10 Binding of phalloidin to
actin ilaments (Problem 16–27).
(A) The actin cytoskeleton stained with
luorescent phalloidin. (B) An actin
ilament bound by gold-tagged phalloidin
at low contrast. (C) The same actin
ilament as in (B), but at high contrast.
Bright bands mark the positions of six
gold particles.
20 μm
(Figure 16–10A). If phalloidin is attached to a gold particle instead, its
binding to actin ilaments can be examined at high resolution by scanning transmission electron microscopy. Figure 16–10B shows a micrograph of an actin ilament with bound phalloidin and Figure 16–10C
shows the same picture with the contrast adjusted so that only the points
of highest intensity (the gold particles) are visible. Does phalloidin bind
to every actin subunit? How can you tell?
Isolated bundles of actin ilaments from the acrosomal processes of
Limulus polyphemus (horseshoe crab) sperm have readily distinguishable plus ends (tapered) and minus ends (blunt). Assembly at the ends
of such bundles was used to determine the mechanism of action of phalloidin, which has a marked efect on actin assembly. When phalloidin is
mixed with actin in a molar ratio of at least 1:1, the growth rate increases
at both ends, as shown for minus ends in Figure 16–11A. Because growth
rate = kon[actin]initial – kof, these plots have the form y = mx + b, so that the
slope of the line equals kon and the y intercept equals –kof.
A. By analyzing the on and of rates, decide how phalloidin increases the
growth rate of actin ilaments. Explain your reasoning.
B. In Figure 16–11B, actin ilaments grown in the presence or absence of
phalloidin were diluted in the absence of actin monomers and their disassembly was assayed. Do these results conirm or contradict your conclusions from part A? Explain your answer.
C. What is the critical concentration for actin assembly at the minus end
in the absence of phalloidin? What is the critical concentration for actin
assembly at the minus end in the presence of phalloidin?
D. Propose a molecular mechanism for the efects of phalloidin on actin
assembly.
16–28
Swinholide A is a member of a class of lipophilic compounds termed
macrolides, which include a number of useful antibiotics such as
erythromycin, that are synthesized by Actinomycetes. Swinholide A is
a “twin” molecule, composed of two identical halves (Figure 16–12A).
When added to cells growing in culture, swinholide A disrupts the actin
cytoskeleton. Your advisor has shown conclusively that swinholide A
binds a pair of actin monomers. She suspects that swinholide A causes
actin ilaments to depolymerize by sequestering actin subunits in a nonfunctional dimeric form and thus accelerating depolymerization through
mass-action efects. She wants you to test this hypothesis.
You prepare actin ilaments tagged with a probe that luoresces
intensely in the ilament but much less so in the free subunits (or swinholide-bound subunits). his allows you to follow depolymerization readily and rapidly as a loss of luorescence. Just as your advisor predicted,
growth rate
(molecules per second)
30
+ phalloidin
15
– phalloidin
0
–15
0
1
2
actin (µM)
3
(B) DISASSEMBLY
percentage of length
before dilution
16–29
(A) ASSEMBLY
45
+ phalloidin
100
– phalloidin
50
0
0
20
10
time after dilution (minutes)
Figure 16–11 Effects of phalloidin on
actin ilaments (Problem 16–28).
(A) Growth rates at the minus ends of
acrosomal bundles in the presence and
absence of phalloidin. (B) Disassembly
of actin ilaments upon dilution in the
presence and absence of phalloidin.
4
ACTIN AND ACTIN-BINDING PROTEINS
339
(A) SWINHOLIDE A STRUCTURE
(C) SWINHOLIDE A TITRATION
depolymerization rate (nm/sec)
fluorescence (arbitrary units)
(B) DISASSEMBLY ASSAY
0.16
0
0.12
152
0.08
280
1000
0.04
480
0.00
0
10
20
30
40
50
5
4
3
2
1
0
50
0
time (seconds)
100
150
swinholide A (nM)
Figure 16–12 Effects of swinholide A on actin ilaments (Problem 16–29). (A) Structure of swinholide A. The identical halves of
swinholide A are arranged head to tail, so that if the molecule were rotated 180° about the indicated axis (circle with an X in it),
it would superimpose on itself. For this reason it is said to have a twofold axis of symmetry. (B) Time course of actin ilament
depolymerization in the presence and absence of swinholide A. Numbers indicate the concentration of swinholide A (nM) used in
each depolymerization assay. (C) Initial rates of depolymerization as a function of swinholide A concentration.
16–30
16–31
Accessory proteins that regulate the nucleation of actin ilaments promote binding of the Arp2/3 complex to actin ilaments so that most new
ilaments form as branches from existing ones. hese proteins could
stimulate Arp2/3 binding to the sides of existing ilaments or to the plus
end of a growing ilament in a way that does not interfere with growth.
Both possibilities would yield the inal characteristic branched network
of ilaments. To distinguish between these alternatives, you mix the regulatory proteins with the Arp2/3 complex and actin subunits in the presence of actin ilaments that are capped at their plus ends. After a short
incubation you examine the resulting structures by electron microscopy.
How will this experiment distinguish between these alternatives? What
structures would you expect to see according to each model for nucleation by the Arp2/3 complex?
You have two proteins that you suspect cap the ends of actin ilaments. To
determine whether they do and, if so, which protein caps which end, you
measure ilament formation as a function of actin concentration in the
absence of either protein, in the presence of protein 1, and in the presence of protein 2 (Figure 16–13). Which protein caps the plus end and
which caps the minus end? How can you tell? Give examples of proteins
in the cell that you would expect to behave like protein 1 and protein 2.
(A) NO ADDED PROTEIN
mass
filament
monomer
0
0.2 0.4 0.6 0.8
actin concentration ( μM)
(B) PROTEIN 1
mass
filament
monomer
0
0.2 0.4 0.6 0.8
actin concentration (μ M)
(C) PROTEIN 2
monomer
mass
depolymerization increases in the presence of increasing concentrations
of swinholide A (Figure 16–12B). But you notice two features of these
curves that suggest to you that swinholide A may actually sever actin ilaments. One of these features is illustrated in Figure 16–12C, which shows
a nonlinear dependence of the initial rate of depolymerization on the
concentration of swinholide A. A simple mass-action efect—the sequestering of actin monomers by binding to swinholide A—predicts a linear
dependence; however, increasing increments in swinholide A concentration have a progressively greater efect on depolymerization.
A. In Figure 16–12B, why does luorescence reach a plateau value (at about
0.03) instead of decreasing to zero?
B. he other odd feature you noticed about depolymerization in the presence of swinholide A (Figure 16–12B) is that the lines have a “hump” in
them in the irst few seconds (when their luorescence is still above 0.12).
Why does this hump suggest that swinholide A severs actin ilaments?
C. Assuming that swinholide A does sever actin ilaments, is one molecule
enough, or are multiple molecules needed? How do you know?
filament
0
0.2 0.4 0.6 0.8
actin concentration (μ M)
Figure 16–13 Effects of two proteins
on actin polymerization (Problem 16–31).
(A) Polymerization of pure actin.
(B) Actin polymerization in the presence
of protein 1. (C) Actin polymerization in
the presence of protein 2. The mass of
actin, as monomers or ilaments, was
determined at equilibrium.
Chapter 16: The Cytoskeleton
340
(A) TIME-LAPSE MOVIE
(B) EM
10 μm
0
10
20
40
30
50
60
70
Figure 16–14 Movement of a bacterium
through the cytosol on a comet tail of
actin ilaments (Problem 16–32). (A) Timelapse movie. (B) Electron micrograph. The
bacterium is 2 μm in length.
2 μm
time (seconds)
MEDICAL LINKS
he intracellular pathogenic bacterium Listeria monocytogenes propels
itself through the cytosol on a comet tail of actin ilaments (Figure 16–14).
Remarkably, only a single bacterial protein, the transmembrane protein
ActA, is required for this motility. ActA is distributed unequally on the
surface of the bacterium, with maximum concentrations at the pole in
contact with the actin tail. he efects of ActA on actin polymerization
in the presence and absence of the Arp2/3 complex (ARP) are shown in
Figure 16–15A. he irst few seconds of the reactions are shown on an
expanded scale in Figure 16–15B. Polymerization of actin was followed
using pyrene-actin, which exhibits much higher luorescence intensity
when actin is polymerized.
A. What are the efects of ActA and the Arp2/3 complex, separately and
together, on the rate of nucleation of actin ilaments? Explain your
answer.
B. How do you suppose that the polymerization of actin by ActA and the
Arp2/3 complex propels the bacterium across the cell? In the comet
tail of actin ilaments, which ends—plus or minus—do you suppose are
pointed at the bacterium?
16–32
(A) KINETICS OF ACTIN ASSEMBLY
(B) EXPANDED SCALE
10,000
fluorescence
(arbitrary units)
actin + ARP
7500
actin + ARP + ActA
actin + ARP + ActA
5000
actin alone,
actin + ActA
actin alone,
actin + ActA,
actin + ARP
2500
0
0
500
1000
1500
2000
10
time (seconds)
40
70
100
time (seconds)
MYOSIN AND ACTIN
TERMS TO LEARN
myofibril
stress fiber
myosin
DEFINITIONS
Match each deinition below with its term from the list above.
16–33
he motor protein in muscle that generates the force for muscle contraction.
Figure 16–15 Effects of ActA and
the Arp2/3 complex (ARP) on actin
polymerization (Problem 16–32).
(A) kinetics of polymerization of actin
in the presence of ActA and the Arp2/3
complex. (B) kinetics of polymerization
on an expanded scale. In all cases, actin
was present at 2 μM, and ActA and the
Arp2/3 complex were present at 30 nM.
MYOSIN AND ACTIN
16–34
341
Long, highly organized bundle of actin, myosin, and other proteins in
the cytoplasm of muscle cells that contracts by a sliding-ilament mechanism.
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
16–35
Myosin II molecules have two motor domains and a rodlike tail that
allows them to assemble into bipolar ilaments, which are crucial for the
eicient sliding of oppositely oriented actin ilaments past each other.
16–36
Motor neurons trigger action potentials in muscle cell membranes that
open voltage-sensitive Ca2+ channels in T tubules, allowing extracellular
Ca2+ to enter the cytosol, bind to troponin C, and initiate rapid muscle
contraction.
16–37
When activated by Ca2+ binding, troponin C causes troponin I to release
its hold on actin, thereby allowing the tropomyosin molecules to shift
their positions slightly so that the myosin heads can bind to the actin ilaments.
THOUGHT PROBLEMS
16–38
Living systems continually transform chemical free energy into motion.
Muscle contraction, ciliary movement, cytoplasmic streaming, cell division, and active transport are examples of the ability of cells to transduce
chemical free energy into mechanical work. In all these instances, a protein motor harnesses the free energy released in a chemical reaction to
drive an attached molecule (the ligand) in a particular direction. Analysis
of free-energy transduction in favorable biological systems suggests that
a set of general principles governs the process in cells.
1. A cycle of reactions is used to convert chemical free energy into
mechanical work.
2. At some point in the cycle a ligand binds very tightly to the protein
motor.
3. At some point in the cycle the motor undergoes a major conformational change that alters the physical position of the ligand.
4. At some point in the cycle the ainity for the ligand markedly
decreases, allowing the ligand to detach from the motor.
hese principles are illustrated by the two cycles for free-energy
transduction shown in Figure 16–16: (1) the sliding of actin and myosin
ilaments against each other and (2) the active transport of Ca2+ from
(A) SLIDING FILAMENT
(B) ACTIVE TRANSPORT
actin
Ca2+
Pi
ADP
ATP
ADP
ATP ADP
INSIDE
Ca
myosin
ATP
Pi
Pi
Ca
ADP
Pi
ATP
Pi
Pi
Ca
OUTSIDE
Ca2+
Figure 16–16 Transduction of chemical free energy into mechanical work (Problem 16–38). (A) Sliding of actin ilaments relative to
myosin ilaments. (B) Active transport of Ca2+ from the inside to the outside of the cell. In both cycles, arrows are drawn in only one
direction to emphasize their normal operation. The phosphorylation and dephosphorylation steps in the active transport cycle are
catalyzed by enzymes that are not shown in the diagram.
Chapter 16: The Cytoskeleton
342
inside the cell, where its concentration is low, to the cell exterior, where
its concentration is high. An examination of these cycles underscores the
principles of free-energy transduction.
A. What is the source of chemical free energy that powers these cycles, and
what is the mechanical work that each cycle accomplishes?
B. What is the ligand that is bound tightly and then released in each of
the cycles? Indicate the points in each cycle where the ligand is bound
tightly.
C. Identify the conformational changes in the protein motor that constitute
the “power stroke” and “return stroke” of each cycle.
Which one of the following changes takes place when a skeletal muscle
contracts?
A. Z discs move farther apart.
B. Actin ilaments contract.
C. Myosin ilaments contract.
D. Sarcomeres become shorter.
16–39
16–40
Two electron micrographs of striated muscle in longitudinal section are
shown in Figure 16–17. he sarcomeres in these micrographs are in two
diferent states of contraction.
(A)
(B)
1 μm
1 μm
Figure 16–17 Two electron micrographs
of striated muscle in longitudinal section
(Problem 16–40). The micrograph in
(B) is a much lighter exposure than the
one in (A). At the same exposure, the
entire space between the thin dark lines
in (B) would be as dark as the fat dark
band in (A).
MYOSIN AND ACTIN
343
CAGED ATP
NH2
N
CH3
C
O
O
O
O
P
O–
H
NO2
P
O
O
O–
N
P
O
H 2C
O–
Figure 16–18 Caged ATP (Problem 16–43).
N
N
O
H
H
OH
OH
H
CH3
C
LASER
LIGHT
O
NH2
NO2
O
–O
P
O–
N
O
O
P
O–
O
O
P
N
O
H 2C
O–
N
N
O
H
H
OH
OH
H
ATP
A. Using the micrograph in Figure 16–17A, identify the locations of the following:
1. Dark band
2. Light band
3. Z disc
4. Myosin II ilaments
5. Actin ilaments (show plus and minus ends)
6. α-Actinin
7. Nebulin
8. Titin
B. Locate the same features on the micrograph in Figure 16–17B. Be careful!
16–41
Troponin molecules are evenly spaced along an actin ilament with one
troponin bound at every seventh actin molecule. How do you suppose
troponin molecules can be positioned this regularly?
16–42
What two major roles does ATP hydrolysis play in muscle contraction?
DATA HANDLING
As a laboratory exercise, you and your classmates are carrying out experiments on isolated muscle ibers using “caged” ATP (Figure 16–18). Since
caged ATP does not bind to muscle components, it can be added to a
muscle iber without stimulating activity. hen, at some later time it can
be split by a lash of laser light to release ATP instantly throughout the
muscle iber.
To begin the experiment, you treat an isolated, striated muscle iber
with glycerol to make it permeable to nucleotides. You then suspend
it in a bufer containing ATP in an apparatus that allows you to measure any tension generated by iber contraction. As illustrated in Figure
16–19, you measure the tension generated after several experimental
manipulations: removal of ATP by dilution, addition of caged ATP, and
activation of caged ATP by laser light. You are somewhat embarrassed
because your results are very diferent from everyone else’s. In checking
over your experimental protocol, you realize that you forgot to add Ca2+
to your bufers. he teaching assistant in charge of your section tells you
that your experiment is actually a good control for the class, but you will
have to answer the following questions to get full credit.
add caged
ATP
tension
16–43
laser
flash
remove
ATP
0
2
4
6
8
10
time (minutes)
Figure 16–19 Tension in a striated muscle
iber as a result of various experimental
manipulations (Problem 16–43).
Chapter 16: The Cytoskeleton
tension (% of maximum)
344
Figure 16–20 Tension as a function
of sarcomere length during isometric
contraction (Problem 16–44).
2.0 2.2
1.6
100
II
III
75
50
1.3
0
I
IV
25
1
3.6
2
3
sarcomere length (μm)
4
A. Why did the ATP in the suspension bufer not cause the muscle iber to
contract?
B. Why did the subsequent removal of ATP generate tension? Why did tension develop so gradually? (If our muscles normally took a full minute to
contract, we would move very slowly.)
C. Why did laser illumination of a iber containing caged ATP lead to relaxation?
16–44
Detailed measurements of sarcomere length and tension during isometric contraction in striated muscle provided crucial early support for
the sliding-ilament model of muscle contraction. Based on your understanding of the sliding-ilament model and the structure of a sarcomere,
propose a molecular explanation for the relationship of tension to sarcomere length in the portions of Figure 16–20 marked I, II, III, and IV. (In
this muscle, the length of the myosin ilament is 1.6 μm, and the lengths
of the actin thin ilaments that project from the Z discs are 1.0 μm.)
MICROTUBULES
TERMS TO LEARN
axoneme
centriole
centrosome
cilium
dynamic instability
dynein
flagellum
γ-tubulin ring complex (γ-TuRC)
kinesin
kinesin-1
microtubule-associated protein (MAP)
microtubule-organizing center (MTOC)
tubulin
DEFINITIONS
Match each deinition below with its term from the list above.
16–45
he property of sudden conversion from growth to shrinkage, and vice
versa, in a protein ilament such as a microtubule or an actin ilament.
16–46
Centrally located organelle of animal cells that is the primary microtubule-organizing center and acts as the spindle pole during mitosis.
16–47
Protein assembly containing a special form of tubulin, along with other
proteins, that is an eicient nucleator of microtubule growth.
16–48
Short cylindrical array of microtubules, a pair of which are embedded in
the major microtubule-organizing center of an animal cell.
16–49
A member of the family of motor proteins that move along microtubules
by walking toward the minus end.
16–50
A motor protein that moves along microtubules by walking toward the
plus end.
MICROTUBULES
16–51
Bundle of microtubules and associated proteins that forms the core of a
cilium or lagellum in a eukaryotic cell and is responsible for their movements.
16–52
Long, hairlike protrusion from the surface of a eukaryotic cell whose
undulations drive the cell through a liquid medium.
345
seam
+ end
TRUE/FALSE
Decide whether each of these statements is true or false, and then explain why.
16–53
he structural polarity of all microtubules is such that α-tubulin is
exposed at one end and β-tubulin is exposed at the opposite end.
16–54
he role of GTP hydrolysis in tubulin polymerization is similar to the
role of ATP hydrolysis in actin polymerization: both serve to weaken the
bonds in the polymer and thereby promote depolymerization.
16–55
All microtubule-organizing centers contain centrioles that help nucleate
microtubule polymerization.
16–56
In most animal cells, minus-end directed microtubule motors deliver
their cargo to the periphery of the cell, whereas plus-end directed microtubule motors deliver their cargo to the interior of the cell.
THOUGHT PROBLEMS
16–57
Why do you suppose it is much easier to add tubulin to existing microtubules than to start a new microtubule from scratch?
16–58
In a 13-ilament microtubule, the majority of lateral interactions are
between like subunits, with α-tubulin binding to α-tubulin, and β-tubulin
binding to β-tubulin. Between the irst and thirteenth protoilaments,
however, there is a seam at which α-tubulin interacts with β-tubulin
(Figure 16–21). Are these heterotypic interactions (α with β) likely to be
stronger than, weaker than, or the same strength as homotypic interactions (α with α, or β with β)? Explain your reasoning.
16–59
he microtubules in Figure 16–22A were obtained from a population
that was growing rapidly, whereas the one in Figure 16–22B came from
microtubules undergoing catastrophic shrinkage. Comment on any differences between the two images and suggest likely explanations for
those you observe.
Dynamic instability causes microtubules either to grow or to shrink rapidly. Consider an individual microtubule that is in its shrinking phase.
A. What must happen at the end of the microtubule in order for it to stop
shrinking and start growing?
B. How would an increase in the tubulin concentration afect this switch
from shrinking to growing?
C. What would happen if GDP, but no GTP, were present in the solution?
D. What would happen if the solution contained an analog of GTP that
could not be hydrolyzed?
8 nm
β
α
– end
tubulin
protofilament
microtubule
Figure 16–21 Structure of a
13-protoilament microtubule, showing
the seam between the irst and thirteenth
protoilaments (Problem 16–58).
16–60
16–61
he β-tubulin subunit of an αβ-tubulin dimer retains its bound GTP for
a short time after it has been added to a microtubule, yielding a GTP
cap whose size depends on the relative rates of polymerization and GTP
hydrolysis. A simple notion about microtubule growth dynamics is that
the ends with GTP caps grow, whereas ends without GTP caps shrink. To
test this idea, you allow microtubules to form under conditions where
you can watch individual microtubules. You then sever one microtubule
in the middle using a laser beam. Would you expect the newly exposed
plus and minus ends to grow or to shrink? Explain your answer.
(A) GROWING
(B) SHRINKING
Figure 16–22 Electron microscopic
analysis of microtubule dynamics
(Problem 16–59). (A) Rapidly growing
microtubules. (B) Catastrophically
shrinking microtubule.
Chapter 16: The Cytoskeleton
346
LINEAR GROWTH
LATERAL ASSOCIATION
16–62
he drugs Taxol®, extracted from the bark of yew trees, and colchicine, an
alkaloid from autumn crocus, have opposite efects. Taxol binds tightly to
microtubules and stabilizes them. When added to cells, it causes much
of the free tubulin to assemble into microtubules. In contrast, colchicine
prevents microtubule formation. Taxol and colchicine are equally toxic
to dividing cells, and both are used as anticancer drugs. Based on your
knowledge of microtubule dynamics, explain why these drugs are toxic
to dividing cells despite their opposite modes of action.
16–63
A solution of pure αβ-tubulin dimers is thought to nucleate microtubules by forming a linear protoilament about seven dimers in length.
At that point, the probabilities that the next αβ-dimer will bind laterally
or to the end of the protoilament are about equal. he critical event for
microtubule formation is thought to be the irst lateral association (Figure 16–23). How does lateral association promote the subsequent rapid
formation of a microtubule?
16–64
How does a centrosome “know” when it has found the center of the cell?
16–65
How are γ-TuRC and the Arp2/3 complex similar, and how are they different?
16–66
When cells enter mitosis, their existing array of cytoplasmic microtubules has to be rapidly broken down and replaced with the mitotic spindle, which pulls the chromosomes into the daughter cells. he enzyme
katanin, named after Japanese samurai swords, is activated during the
onset of mitosis and cleaves microtubules into short pieces. What do you
suppose is the fate of the microtubule fragments created by katanin?
16–67
Kinesin-1 motors are highly processive, moving long distances on microtubule tracks without dissociating. By contrast, myosin II motors in skeletal muscle do not move processively; they take only one or a few steps
before letting go. How are these diferent degrees of processivity adapted
to the biological functions of kinesin-1 and myosin II?
Figure 16–23 Model for microtubule
nucleation by pure αβ-tubulin dimers
(Problem 16–63).
8
7
1
6
16–68 An electron micrograph of a cross section through a lagellum is shown in
Figure 16–24.
A. Assign the following components to the indicated positions on the igure.
A microtubule
B microtubule
Outer dynein arm
Inner dynein arm
Inner sheath
Nexin
Radial spoke
Singlet microtubule
B. Which of the above structures are composed of tubulin?
2
3
4
5
100 nm
Figure 16–24 Electron micrograph of
a cross section through a lagellum of
Chlamydomonas reinhardtii (Problem
16–68).
MICROTUBULES
16–69
16–70
he sliding-microtubule mechanism for ciliary bending is undoubtedly
correct. he consequences of sliding are straightforward when a pair of
outer doublets is considered in isolation. he dynein arms are arranged
so that, when activated, they push their neighboring outer doublet outward toward the tip of the cilium. If the pair of outer doublets is linked
together by nexin molecules, they will bend so that the one that has
been pushed toward the tip will deine the inside of the curve (see Figure
16–28A). It is confusing, however, to think about sliding in the circular
array of outer doublets in the axoneme. If all the dynein arms in a circular
array were equally active, there could be no signiicant relative motion.
(he situation is equivalent to a circle of strongmen, each trying to lift his
neighbor of the ground; if they all succeeded, the group would levitate.)
Devise a pattern of dynein activity (consistent with axoneme structure and the directional pushing of dynein) that could account for bending of the axoneme in one direction. How would this pattern change for
bending in the opposite direction?
347
hooks
microtubules
Figure 16–25 Tubulin-decorated
microtubules in a cross section through
a nerve axon (Problem 16–70). The hooks
represent the tubulin decoration.
In addition to conducting impulses in both directions, nerve axons carry
vesicles to and from the cell body along microtubule tracks. Do outbound
vesicles move along microtubules that are oriented in one direction and
incoming vesicles move along oppositely oriented microtubules? Or are
microtubules all oriented in the same direction, with diferent motor
proteins providing the directionality?
To distinguish between these possibilities, you prepare a cross section through a nerve axon and decorate the microtubules with tubulin,
which binds to the tubulin subunits of the microtubule to form hooks.
he decorated microtubules are illustrated in Figure 16–25. Do all the
microtubules run in the same direction or not? How can you tell?
CALCULATIONS
he function of microtubules depends on their speciic spatial organization within the cell. How are speciic arrangements created, and what
determines the formation and disappearance of individual microtubules?
To address these questions, investigators have studied the in
vitro assembly of αβ-tubulin dimers into microtubules. Below 15 μM
αβ-tubulin, no microtubules are formed; above 15 μM, microtubules
form readily (Figure 16–26A). If centrosomes are added to the solution of
tubulin, microtubules begin to form at less than 5 μM (Figure 16–26B).
(Diferent assays were used in the two experiments—total weight of
microtubules in Figure 16–26A and the average number of microtubules
per centrosome in Figure 16–26B—but the lowering of the critical concentration for microtubule assembly in the presence of centrosomes is
independent of the method of assay.)
A. Why do you think that the concentration at which microtubules begin to
form (the critical concentration) is diferent in the two experiments?
B. Why do you think that the plot in Figure 16–26A increases linearly with
increasing tubulin concentration above 15 μM, whereas the plot in Figure 16–26B reaches a plateau at about 25 μM?
C. he concentration of αβ-tubulin dimers (the subunits for assembly) in
a typical cell is 1 mg/mL and the molecular weight of a tubulin dimer is
110,000. What is the molar concentration of tubulin dimers in cells? How
does the cellular concentration compare with the critical concentrations
in the two experiments in Figure 16–26? What are the implications for the
assembly of microtubules in cells?
16–72
(A) NO CENTROSOMES
microtubule mass
At 1.4 mg/mL pure tubulin, microtubules grow at a rate of about 2 μm/
min. At this growth rate, how many αβ-tubulin dimers (8 nm in length)
are added to the ends of a microtubule each second?
40
30
20
10
0
0
10
20
30
40
50
(B) WITH CENTROSOMES
microtubule number
16–71
60
40
20
0
0
10
20
30
40
50
αβ -tubulin (µM)
Figure 16–26 Analysis of microtubule
assembly (Problem 16–72). (A) Mass of
microtubules assembled in the absence
of centrosomes as a function of tubulin
concentration. Tubulin assembly into
microtubules was measured by the
increase in solution turbidity. (B) Average
number of microtubules per centrosome
as a function of tubulin concentration.
Concentrations refer to αβ-tubulin dimers,
which are the subunits of assembly.
Chapter 16: The Cytoskeleton
348
(A) POLYMERIZATION KINETICS
(B) CRITICAL CONCENTRATION
polymerized tubulin
(µM)
tubulin assembly
0.3
+ γ-tubulin
10
0.2
– γ-tubulin
0.1
0
0
30
time (minutes)
60
+ γ-tubulin
5
Figure 16–27 Effects of γ-tubulin on
microtubule polymerization (Problem 16–
73). (A) kinetics of polymerization in the
presence and absence of γ-tubulin.
(B) The critical concentration for
microtubule assembly in the presence of
0.6 nM γ-tubulin and in its absence.
– γ-tubulin
0
0
5
10
αβ-tubulin (µM)
he γ-tubulin ring complex (γ-TuRC), which nucleates microtubule
assembly in cells, includes γ-tubulin and several accessory proteins.
To get at its mechanism of nucleation, you have prepared monomeric
γ-tubulin by in vitro translation and puriication. You measure the efect
of adding monomeric γ-tubulin to a solution of αβ-tubulin dimers, as
shown in Figure 16–27.
A. In the presence of monomeric γ-tubulin, the lag time for assembly of
microtubules decreases, and assembly occurs more rapidly (Figure
16–27A). How would you account for these two efects of γ-tubulin?
B. he critical concentration of αβ-tubulin needed for the assembly of
microtubules is reduced from about 3.2 μM in the absence of γ-tubulin
to about 1.7 μM in its presence (Figure 16–27B). How do you suppose
γ-tubulin lowers the critical concentration? How does this account for
the greater extent of polymerization in Figure 16–27A? (hink about the
end—plus or minus—at which polymerization occurs in the presence of
γ-tubulin.)
16–73
16–74
Using the equation for difusion given in Problem 16–10, calculate the
average time it would take for a vesicle to difuse to the end of an axon 10
cm in length. he difusion coeicient of a typical vesicle is 5 × 10–8 cm2/
sec.
16–75
A mitochondrion 1 μm long can travel the 1 meter length of the axon
from the spinal cord to the big toe in a day. he Olympic men’s freestyle
swimming record for 200 meters is 1.75 minutes. In terms of body lengths
per day, who is moving faster: the mitochondrion or the Olympic record
holder? (Assume that the swimmer is 2 meters tall.)
During the lagellar beat cycle in Chlamydomonas, the bent segment of
the lagellum extends roughly through half the circumference of a circle
(Figure 16–28).
A. How much sliding of microtubule doublets against one another is
required to account for the observed bending of the lagellum into a
semicircle? Calculate how much farther the doublet on the inside of the
semicircle protrudes beyond the doublet on the outside of the semicircle
at the tip of the lagellum (Figure 16–28A). he width of a lagellum is
180 nm.
B. he elastic protein molecule (nexin) that links adjacent outer doublets
must stretch to accommodate the bending of a lagellum into a semicircle. If the length of an unstretched nexin molecule at the base of a lagellum is 30 nm, what is the length of a stretched molecule at the tip of a
lagellum (Figure 16–28B)? Adjacent doublets are 30 nm apart.
(A)
outside outer doublet
r
r + 180
base of
axoneme
inside
outer
doublet
(B)
16–76
DATA HANDLING
16–77
he orientation of the αβ-tubulin dimer in a microtubule was determined in several ways. GTP-coated luorescent beads, for example, were
found to bind exclusively at the plus ends of microtubules. By contrast,
gold beads coated with antibodies speciic for a peptide of α-tubulin
r
r + 30
unstretched
nexin
stretched
nexin
Figure 16–28 Flagella bent into halfcircles (Problem 16–76).
(A) Representation showing the
“inside” and “outside” doublets, which
are 180 nm apart. (B) Representation
showing adjacent doublets, which are
30 nm apart, and the nexin molecules
that link them.
MICROTUBULES
349
Figure 16–29 Analysis of growth kinetics
of individual microtubules (Problem
16–78). (A) Changes in length at the
plus ends. Results from two individual
microtubules are indicated by different
shades of red. (B) Changes in length
at the minus ends. Results from two
individual microtubules are indicated by
different shades of blue.
(A) PLUS ENDS
length ( µm)
8
a
4
0
b
0
5
10
time (minutes)
15
20
length ( µm)
(B) MINUS ENDS
4
b
2
0
0
5
10
15
20
time (minutes)
bound exclusively at the minus end. How do these observations deine
the orientation of the αβ-tubulin dimer in the microtubule? Which tubulin subunit, α or β, is at which end? Explain your reasoning.
he complex kinetics of microtubule assembly make it hard to predict the
behavior of individual microtubules. Some microtubules in a population
can grow, even as the majority shrink to nothing. One simple hypothesis
proposed to explain this behavior is that a growing end is protected from
disassembly by a GTP cap and that a faster-growing end has a longer
GTP cap. Real-time video observations of changes in length with time are
shown for two individual microtubules in Figure 16–29. Measurements
of their rates of growth and shrinkage show that the plus end of each
microtubule grows three times faster than the minus end, and shrinks at
half the rate.
A. Are changes in length at the two ends of a microtubule dependent or
independent of one another? How can you tell?
B. What does the GTP-cap hypothesis predict about the rate of switching
between growing and shrinking states at the fast-growing end relative to
the slow-growing end? Does the outcome of this experiment support the
GTP-cap hypothesis?
C. What do you suppose would happen if centrosomes were used to nucleate
growth? What would happen if microtubule-associated proteins (MAPs)
were included?
16–78
Comparisons of microtubule behavior between species point to diferences that raise questions about the biological importance of dynamic
instability. Notothenioid ish, for example, which live in the Southern
Ocean at a constant temperature of –1.8°C, have remarkably stable microtubules compared with warm-blooded vertebrates such as the cow. his
is an essential modiication for notothenioid ish because normal microtubules disassemble completely into αβ-tubulin dimers at 0°C. Measurements on individual microtubules in solutions of pure tubulin show
that notothenioid ish microtubules grow at a much slower rate, shrink
at a much slower rate, and only rarely switch from growth to shrinkage
(catastrophe) or from shrinkage to growth (rescue) (Table 16–1).
A. he amino acid sequences of the α- and β-tubulin subunits from notothenioid ish difer from those of the cow at positions and in ways that
might reasonably be expected to stabilize the microtubule, in accord with
16–79
Chapter 16: The Cytoskeleton
350
TABLE 16–1 Properties of individual microtubules in notothenioid fish and the
domestic cow (Problem 16–79).
Growth
rate
(μm/min)
Shrinkage
rate
(μm/min)
Catastrophe
frequency
(min–1)
Rescue
frequency
(min–1)
Notothenioid fish
0.27
0.8
0.008
<0.0004
Domestic cow
2.18
61.2
0.52
3.1
Microtubules
Multiple individual microtubules were observed by video microscopy near the body temperature
for each species: 5°C for fish and 37°C for cow. Average growth rates were calculated for
growing microtubules, and average shrinkage rates were calculated for shrinking microtubules.
Changes from growth to shrinkage (catastrophe) and from shrinkage to growth (rescue) were
averaged over the observation period and expressed as frequency of events per minute.
the data in Table 16–1. Would you expect these changes to strengthen
the interactions between the α- and β-tubulin subunits in the αβ-dimer,
between adjacent dimers in the protoilament, or between tubulin subunits in adjacent protoilaments? Explain your reasoning.
B. Dynamic instability is thought to play a fundamental role in the rapid
microtubule rearrangements that occur in cells. How do you suppose
cells in these notothenioid ishes manage to alter their microtubule
architecture quickly enough to accomplish essential cell functions? Or
do you suppose that these cells exist with a stable microtubule cytoskeleton that only slowly rearranges itself?
16–80
A standard puriication scheme for tubulin is to prepare a cell extract,
chill it to 0°C, spin it at high speed, and save the supernatant. he supernatant is then warmed to 37°C and incubated in the presence of GTP. he
mixture is then spun at high speed and the pellet is saved and redissolved.
hen the cycle is repeated: chill the dissolved pellet, spin at high speed,
save the supernatant, incubate with GTP at 37°C, spin at high speed, save
the pellet. After a few cycles one obtains a pure preparation of tubulin.
Explain how this procedure yields pure tubulin.
16–81
In addition to centrosomes, lagellar axonemes also can serve as nucleation sites for microtubule assembly. he following experiment was
designed to determine whether these two structures nucleate microtubule growth by binding to the plus end or to the minus end of the nascent
microtubule. Flagellar axonemes were included as a control since their
plus and minus ends can be distinguished. Centrosomes and lagellar
axonemes were incubated briely in unlabeled tubulin to nucleate microtubule growth. A high concentration of biotin-labeled tubulin was then
added and the incubation was continued for 10 minutes. At that point the
preparations were ixed and the biotin-labeled segments were visualized
by adding luorescein-labeled antibodies speciic for biotin. he lengths
of the biotin-labeled segments were measured and plotted as shown in
Figure 16–30.
(A) AXONEMES
(B) CENTROSOMES
60
80
microtubule
number
minus end
60
40
40
plus end
20
0
20
0
5
10
length (µm)
15
0
0
5
10
length (µm)
15
Figure 16–30 length distributions of
microtubules (Problem 16–81).
(A) Nucleation by lagellar axonemes,
whose plus and minus ends can
be distinguished. (B) Nucleation by
centrosomes.
MICROTUBULES
351
(C) MINUS ENDS
(B) PLUS ENDS
(A) MICROGRAPHS
60
% total microtubules
60
– γ-TuRC
50
– γ-TuRC
50
40
40
30
30
20
20
10
10
0
0
0 3 6 9 12 15 18 21
0 3 6 9 12 15 18 21
60
% total microtubules
60
+ γ-TuRC
50
– γ-TuRC
+ γ-TuRC
50
40
40
30
30
20
20
10
10
Figure 16–31 Effects of γ-TuRC on
microtubule assembly (Problem 16–82).
(A) Example of microtubules grown in
the presence and absence of γ-TuRC.
Scale bar is 10 μm. (B) Distribution of the
lengths of bright segments at microtubule
plus ends in the presence and absence
of γ-TuRC. (C) Distributions of the lengths
of bright segments at microtubule minus
ends in the presence and absence of
γ-TuRC. In (B) and (C), only microtubules
with a deined dim segment and one
or two bright terminal segments were
counted.
+ γ-TuRC
0
0
0 3 6 9 12 15 18 21
0 3 6 9 12 15 18 21
microtubule length (µm)
microtubule length (µm)
A. Which end of a newly assembled microtubule is attached to the plus end
of the lagellar axoneme?
B. Which end of a microtubule assembled on a lagellar axoneme grows
faster?
C. Which end of an assembled microtubule is attached to a centrosome?
Explain your reasoning.
16–82
In the paper that deined the γ-tubulin ring complex (γ-TuRC), the
authors puriied the complex from Xenopus oocytes and showed that it
dramatically stimulated the nucleation of microtubules. To determine
whether nucleation occurred at plus ends or at minus ends, they polymerized microtubules in two steps. In the irst step, microtubules nucleated with or without γ-TuRC were allowed to form in the presence of a
small amount of αβ-tubulin containing a low proportion of rhodaminelabeled αβ-tubulin, which makes the microtubules luoresce dimly. In
the second step, these microtubules were allowed to extend at both ends
in the presence of a higher proportion of rhodamine-tagged tubulin to
label the ends brightly. he longer bright segment identiies the plus end,
and the shorter segment the minus end (Figure 16–31A). Measurements
of the lengths of a large number of bright segments in individual microtubules yielded the data in Figure 16–31B and C. At which end of the
microtubule is γ-TuRC when it nucleates? Explain your reasoning.
16–83
A useful technique for studying a microtubule motor is to attach the
motor proteins by their tails to a glass coverslip (the tails stick avidly to a
clean glass surface) and then to allow microtubules to settle onto them.
In the light microscope, the microtubules can be seen to move over the
surface of the coverslip as the heads of the motors propel them (Figure
16–32).
5 µM
0 sec
24 sec
48 sec
72 sec
Figure 16–32 Movement of microtubules
on a bed of microtubule motor molecules
(Problem 16–83). Red arrows mark the
movement of a microtubule with a gold
bead attached via antibodies to the
minus end of a microtubule; white arrows
mark the movement of a microtubule
without an attached bead. Pictures
were taken using video-enhanced
differential-interference-contrast (VE-DIC)
microscopy.
Figure 16–33 Microtubules with α-tubulin
antibody-coated gold beads attached to
one end (Problem 16–84). At the vertical
line a section of each microtubule has
been removed so that the two ends can
be displayed side by side.
landing rate
(microtubules/sec/mm2 )
Chapter 16: The Cytoskeleton
352
100
16–84
In Problem 16–77, the orientation of the αβ-tubulin dimer in the microtubule was determined by showing that α-tubulin antibody-coated gold
beads bound to the minus end. he electron micrographs, however, just
showed microtubules with beads at one end (Figure 16–33). How do you
suppose the investigators knew which end was which? Design an experiment to determine the orientation of microtubules labeled at one end
with a gold bead.
Kinesin carries vesicles for long distances along microtubule tracks in
the cell. Are the two motor domains of a kinesin molecule essential to
accomplish this task, or could a one-headed motor protein function just
as well? Using recombinant DNA techniques, a version of kinesin was
prepared that was identical to normal kinesin except that one motor
domain was absent. Wild-type kinesin with two motor domains and
recombinant kinesin with one were attached to coverslips at a variety
of densities and the rate at which microtubules were bound and moved
(collectively, termed the landing rate) was measured (Figure 16–34).
A. Why do you suppose that the curves at low motor densities are so diferent?
B. What do these experiments say about the design of the kinesin motor: are
two heads required for vesicle transport, or is only one needed? Explain
your reasoning.
16–85
16–86
he movements of single motor-protein molecules can be analyzed
directly. Using polarized laser light, it is possible to create interference
patterns that exert a centrally directed force, ranging from zero at the
center to a few piconewtons at the periphery (about 200 nm from the
center). Individual molecules that enter the interference pattern are rapidly pushed to the center, allowing them to be captured and moved at the
experimenter’s discretion.
Using such “optical tweezers,” single kinesin molecules can be positioned on a microtubule that is ixed to a coverslip. Although a single
kinesin molecule cannot be seen optically, it can be tagged with a silica
bead and tracked indirectly by following the bead (Figure 16–35A). In
the absence of ATP, the kinesin molecule remains at the center of the
interference pattern, but with ATP it moves toward the plus end of the
microtubule. As kinesin moves along the microtubule, it encounters the
force of the interference pattern, which simulates the load kinesin carries
during its actual function in the cell. Moreover, the pressure against the
silica bead counters the efects of Brownian (thermal) motion, so that the
position of the bead more accurately relects the position of the kinesin
molecule on the microtubule.
1 head
1
0.1
A. Since the motor proteins attach in random orientations to the coverslip,
how can they generate coordinated movement of individual microtubules, rather than engaging in a tug-of-war?
B. In which direction will microtubules crawl on a bed of dynein motor
molecules (that is, will they move plus-end irst or minus-end irst)?
C. In the experiment shown in Figure 16–32, some of the microtubules
were marked by gold beads that were bound by minus-end-speciic antibodies. Is the motor protein on the coverslip a plus-end or minus-end
directed motor? How can you tell?
2 heads
10
1
10
100
1000
molecules/µm2
Figure 16–34 landing rates—binding and
moving—of microtubules as a function
of motor protein density (Problem 16–85).
Results with wild-type kinesin are shown
as green square